A Geometric Approach to Differential Forms

DAVID BACHMAN

Birkhäuser

Table of Contents

Title PageCopyright Page
Dedication
Preface
Guide to the Reader
Chapter 1 - Multivariable Calculus
1.1 Vectors
1.2 Functions of multiple variables
1.3 Multiple integrals
1.4 Partial derivatives
1.5 Gradients
Chapter 2 - Parameterizations
2.1 Parameterized curves in
2.2 Cylindrical and spherical coordinates
2.3 Parameterized surfaces in
2.4 Parameterized curves in
2.5 Parameterized regions in and
Chapter 3 - Introduction to Forms
3.1 So what is a differential form?
3.2 Generalizing the integral
3.3 Interlude: a review of single variable integration
3.4 What went wrong?
3.5 What about surfaces?
Chapter 4 - Forms
4.1 Coordinates for vectors
4.21 -forms
4.3 Multiplying 1-forms
4.4 2-forms on T _(optional)
4.5 2-forms and 3-forms on T (optional)
4.6 n-forms
4.7 Algebraic computation of products
Chapter 5 - Differential Forms
5.1 Families of forms
5.2 Integrating differential 2-forms
5.3 Orientations
5.4 Integrating 1 -forms on R R RRR
5.5 Integrating n-forms on R
5.6 The change of variables formula
5.7 Summary: How to integrate a differential form
Chapter 6 - Differentiation of Forms
6.1 The derivative of a differential 1-form
6.2 Derivatives of n n nnn-forms
6.3 Interlude: 0 -forms
6.4 Algebraic computation of derivatives
6.5 Antiderivatives
Chapter 7 -Stokes' Theorem
7.1 Cells and chains
7.2 The generalized Stokes' Theorem
7.3 Vector calculus and the many faces of the generalized Stokes' Theorem
Chapter 8 - Applications
8.1 Maxwell's equations

8.2 Foliations and contact structures

8.3 How not to visualize a differential 1-form
Chapter 9 - Manifolds
9.1 Pull-backs
9.2 Forms on subsets of R R RRR
9.3 Forms on parameterized subsets
9.4 Forms on quotients of R (optional).
9.5 Defining manifolds
9.6 Differential forms on manifolds
9.7 Application: DeRham cohomology.
A - Non-linear forms
A. 1 Surface area
A. 2 Arc length
References
Index
Solutions

David Bachman

A Geometric Approach to Differential Forms

BirkhäuserBoston • Basel • Berlin

David Bachman
Pitzer College
Department of Mathematics
Claremont, CA 91711
U.S.A.
Cover design by Mary Burgess.
Mathematics Subject Classification (2000): Primary: 58A10; Secondary: 26B12, 26B20
Library of Congress Control Number: 2006930548
ISBN-10: 0-8176-4499-7 eISBN: 0-8176-4520-9
ISBN-13: 978-0-8176-4499-4
Printed on acid-free paper.
© 2006 Birkhäuser BostonBirkhduser (2) ( 8 ) ( 8 ) ^((8)){ }^{(8)}(8)
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To Sebastian and Simon

Preface

The present work is not meant to contain any new material about differential forms. There are many good books out there which give complete treatments of the subject. Rather, the goal here is to make the topic of differential forms accessible to the sophomore level undergraduate, while still providing material that will be of interest to more advanced students.
There are three tracks through this text. The first is a course in Multivariable Calculus, suitable for the third semester in a standard calculus sequence. The second track is a sophomore level Vector Calculus class. The last track is for advanced undergraduates, or even beginning graduate students. At many institutions, a course in linear algebra is not a prerequisite for either multivariable calculus or vector calculus. Consequently, this book has been written so that the earlier chapters do not require many concepts from linear algebra. What little is needed is covered in the first section.
The book begins with basic concepts from multivariable calculus such as partial derivatives, gradients and multiple integrals. All of these topics are introduced in an informal, pictorial way to quickly get students to the point where they can do basic calculations and understand what they mean. The second chapter focuses on parameterizations of curves, surfaces and three-dimensional regions. We spend considerable time here developing tools which will help students find parameterizations on their own, as this is a common stumbling block.
Chapter 3 is purely motivational. It is included to help students understand why differential forms arise naturally when integrating over parameterized domains.
The heart of this text is Chapters 4 through 7. In these chapters, the entire machinery of differential forms is developed from a geometric standpoint. New ideas are always introduced with a picture. Verbal descriptions of geometric actions are set out in boxes.
Chapter 7 focuses on the development of the generalized Stokes' Theorem. This is really the centerpiece of the text. Everything that precedes it is there for the sole purpose of its development. Everything that follows is an application. The equation is simple:
C ω = C d ω . C ω = C d ω . int_(del C)omega=int_(C)d omega.\int_{\partial C} \omega=\int_{C} d \omega .Cω=Cdω.
Yet it implies, for example, all integral theorems of classical vector analysis. Its simplicity is precisely why it is easier for students to understand and remember than these classical results.
Chapter 7 concludes with a discussion on how to recover all of vector calculus from the generalized Stokes' Theorem. By the time students get through this they tend to be more proficient at vector integration than after traditional classes in vector calculus. Perhaps this will allay some of the concerns many will have in adopting this textbook for traditional classes.
Chapter 8 contains further applications of differential forms. These include Maxwell's equations and an introduction to the theory of foliations and contact structures. This material should be accessible to anyone who has worked through Chapter 7.
Chapter 9 is intended for advanced undergraduate and beginning graduate students. It focuses on generalizing the theory of differential forms to the setting of abstract manifolds. The final section contains a brief introduction to DeRham cohomology.
We now describe the three primary tracks through this text.
Track 1. Multivariable Calculus (Calculus III). For such a course, one should focus on the definitions of n n nnn-forms on m m mmm, where n n nnn and m m mmm are at most 3. The following Chapters/Sections are suggested:
  • Chapter 1, perhaps supplementing Section 1.5 with additional material on max/min problems,
  • Chapter 2,
  • Chapter 4, excluding Sections 4.4 and 4.5 due to time constraints,
  • Chapters 5-7,
  • Appendix A.
Track 2. Vector Calculus. In this course, one should mention that for n n nnn-forms on R m R m R^(m)\mathbb{R}^{m}Rm the numbers n n nnn and m m mmm could be anything, although in practice it is difficult to work examples when either is bigger than 4. The following Chapters/Sections are suggested:
  • Section 1.1 (unless Linear Algebra is a prerequisite),
  • Chapter 2,
  • Chapter 3 (one lecture),
  • Chapters 4-7,
  • Chapter 8, as time permits.
Track 3. Upper Division Course. Students should have had linear algebra, and perhaps even basic courses in group theory and topology.
  • Chapter 3 (perhaps as a reading assignment),
  • Chapters 4-7 (relatively quickly),
  • Chapters 8 and 9.
The original motivation for this book came from [GP74], the text I learned differential topology from as a graduate student. In that text, differential forms are defined in a highly algebraic manner, which left me craving something more intuitive. In searching for a more geometric interpretation, I came across Chapter 7 of Arnold's text on classical mechanics [Arn97], where there is a wonderful introduction to differential forms given from a geometric viewpoint. In some sense, the present work is an expansion of the presentation given there. Hubbard and Hubbard's text [HHO1] was also a helpful reference during the preparation of this manuscript.
The writing of this book began with a set of lecture notes from an introductory course on differential forms, given at Portland State University, during the summer of 2000. The notes were then revised for subsequent courses on multivariable calculus and vector calculus at California Polytechnic State University, San Luis Obispo and Pitzer College.
I thank several people. First and foremost, I am grateful to all those students who survived the earlier versions of this book. I would also like to thank several of my colleagues for giving me helpful comments. Most notably, Don Hartig, Matthew White and Jim Hoste had several comments after using earlier versions of this text for vector or multivariable calculus courses. John Etnyre and Danny Calegari gave me feedback regarding Chapter 8 and Saul Schleimer suggested Example 27. Other helpful suggestions were provided by Ryan Derby-Talbot. Alvin Bachman suggested some of the formatting of the text. Finally, the idea to write this text came from conversations with Robert Ghrist while I was a graduate student at the University of Texas at Austin.
Claremont, CA
March, 2006
David Bachman

Guide to the Reader

It often seems like there are two types of students of mathematics: those who prefer to learn by studying equations and following derivations, and those who prefer pictures. If you are of the former type, this book is not for you. However, it is the opinion of the author that the topic of differential forms is inherently geometric, and thus should be learned in a visual way. Of course, learning mathematics in this way has serious limitations: how can one visualize a 23 -dimensional manifold? We take the approach that such ideas can usually be built up by analogy to simpler cases. So the first task of the student should be to really understand the simplest case, which CAN often be visualized.
Fig. 0.1. The faces of the n n nnn-dimensional cube come from connecting the faces of two copies of an ( n 1 n 1 n-1n-1n1 )-dimensional cube.
For example, suppose one wants to understand the combinatorics of the n n nnn-dimensional cube. We can visualize a 1-D cube (i.e., an interval), and see just from our mental picture that it has two boundary points. Next, we can visualize a 2-D cube (a square), and see from our picture that this has four intervals on its boundary. Furthermore, we see that we can construct this 2-D cube by taking two parallel copies of our original 1-D cube and connecting the endpoints. Since there are two endpoints, we get two new intervals, in addition to the two we started with (see Fig. 0.1). Now, to construct a 3-D cube, we place two squares parallel to each other, and connect up their edges. Each time we connect an edge of one square to an edge of the other, we get a new square on the boundary of the 3-D cube. Hence, since there were four edges on the boundary of each square, we get four new squares, in addition to the two we started with, making six in all. Now, if the student understands this, then it should not be hard to convince him/her that every time we go up a dimension, the number of lower-
dimensional cubes on the boundary is the same as in the previous dimension, plus two. Finally, from this we can conclude that there are 2 n ( n 1 ) 2 n ( n 1 ) 2n(n-1)2 n(n-1)2n(n1)-dimensional cubes on the boundary of the n n nnn dimensional cube.
Note the strategy in the above example: we understand the "small" cases visually, and use them to generalize to the cases we cannot visualize. This will be our approach in studying differential forms.
Perhaps this goes against some trends in mathematics in the last several hundred years. After all, there were times when people took geometric intuition as proof, and later found that their intuition was wrong. This gave rise to the formalists, who accepted nothing as proof that was not a sequence of formally manipulated logical statements. We do not scoff at this point of view. We make no claim that the above derivation for the number of ( n 1 n 1 n-1n-1n1 )-dimensional cubes on the boundary of an n n nnn-dimensional cube is actually a proof. It is only a convincing argument, that gives enough insight to actually produce a proof. Formally, a proof would still need to be given. Unfortunately, all too often the classical math book begins the subject with the proof, which hides all of the geometric intuition that the above argument leads to.

1

Multivariable Calculus

1.1 Vectors

A vector is a lot like a point in space. The primary difference is that we do not usually think about doing algebra with points, while algebra with vectors is common.
When one switches from talking about points like ( 1 , 2 ) ( 1 , 2 ) (1,2)(1,2)(1,2) to vectors like 1 , 2 1 , 2 (:1,2:)\langle 1,2\rangle1,2, both the language and notation change. We will be very consistent in this text about using parentheses to denote points and brackets to denote vectors. When discussing the point ( 1 , 2 ) ( 1 , 2 ) (1,2)(1,2)(1,2) we say the numbers 1 and 2 are its coordinates. If we are discussing the vector 1 , 2 1 , 2 (:1,2:)\langle 1,2\rangle1,2 then 1 and 2 are its components.
One often visualizes a vector a , b a , b (:a,b:)\langle a, b\ranglea,b as an arrow from the point ( 0, 0 ) to the point ( a , b a , b a,ba, ba,b ). This has some pleasant features. First, it immediately follows from the Pythagorean Theorem that the length of the arrow representing the vector a , b a , b (:a,b:)\langle a, b\ranglea,b is
| a , b | = a 2 + b 2 . | a , b | = a 2 + b 2 . |(:a,b:)|=sqrt(a^(2)+b^(2)).|\langle a, b\rangle|=\sqrt{a^{2}+b^{2}} .|a,b|=a2+b2.
We add vectors just as one would ho e:
a , b + c , d = a + c , b + d . a , b + c , d = a + c , b + d . (:a,b:)+(:c,d:)=(:a+c,b+d:).\langle a, b\rangle+\langle c, d\rangle=\langle a+c, b+d\rangle .a,b+c,d=a+c,b+d.
Geometrically, adding a vector V 1 V 1 V_(1)V_{1}V1 to a vector V 2 V 2 V_(2)V_{2}V2 is equivalent to sliding V 2 V 2 V_(2)V_{2}V2 along V 1 V 1 V_(1)V_{1}V1 until its "tail" is at the "tip" of V 1 V 1 V_(1)V_{1}V1. The vector which represents the sum V 1 + V 2 V 1 + V 2 V_(1)+V_(2)V_{1}+V_{2}V1+V2 is then the one which connects the tail of V 1 V 1 V_(1)V_{1}V1 to the tip of V 2 V 2 V_(2)V_{2}V2. See Figure 1.1.
Multiplication is a bit trickier. The most basic kind of multiplication involves a number and a vector, as follows:
c a , b = c a , c b . c a , b = c a , c b . c(:a,b:)=(:ca,cb:).c\langle a, b\rangle=\langle c a, c b\rangle .ca,b=ca,cb.
1.1. Use similar triangles to show that c a , b c a , b c(:a,b:)c\langle a, b\rangleca,b is a vector that points in the same direction as a , b a , b (:a,b:)\langle a, b\ranglea,b, but has a length that is c c ccc times as large.
Fig. 1.1. Adding vectors.

1.2. Find a vector that points in the same direction as 3 , 4 3 , 4 (:3,4:)\langle 3,4\rangle3,4, but has length one. (Such a vector is called a unit vector.)
To define the product of two vectors the simplest thing to do is to define the product as follows:
a , b c , d = a c , b d . a , b c , d = a c , b d . (:a,b:)(:c,d:)=(:ac,bd:).\langle a, b\rangle\langle c, d\rangle=\langle a c, b d\rangle .a,bc,d=ac,bd.
There is nothing wrong with this, but it does not turn out to be terribly useful. Perhaps the reason is that this definition does not lend itself to a good geometric interpretation.
A more useful way to multiply vectors is called the dot product. The trick with the dot product is to define the product of two vectors to be the number
a , b c , d = a c + b d . a , b c , d = a c + b d . (:a,b:)*(:c,d:)=ac+bd.\langle a, b\rangle \cdot\langle c, d\rangle=a c+b d .a,bc,d=ac+bd.
Fig. 1.2. The dot product of V 1 V 1 V_(1)V_{1}V1 and V 2 V 2 V_(2)V_{2}V2 is L L LLL times the length of V 1 V 1 V_(1)V_{1}V1.
There are two noteworthy things that immediately follow from this definition. First, notice that if V 1 = a , b V 1 = a , b V_(1)=(:a,b:)V_{1}=\langle a, b\rangleV1=a,b, then V 1 V 1 = a 2 + b 2 = V 1 V 1 = a 2 + b 2 = V_(1)*V_(1)=a^(2)+b^(2)=V_{1} \cdot V_{1}=a^{2}+b^{2}=V1V1=a2+b2= | V 1 | 2 V 1 2 |V_(1)|^(2)\left|V_{1}\right|^{2}|V1|2. Second, notice that the slope of the line containing V 1 V 1 V_(1)V_{1}V1 is b / a b / a b//a\mathrm{b} / \mathrm{a}b/a. If V 2 = c , a V 2 = c , a V_(2)=(:c,a:)V_{2}=\langle c, a\rangleV2=c,a is perpendicular to V 1 V 1 V_(1)V_{1}V1 then d / c = a / b d / c = a / b d//c=-a//b\mathrm{d} / \mathrm{c}=-\mathrm{a} / \mathrm{b}d/c=a/b. Cross-multiplying then gives b d = a c b d = a c bd=-acb d=-a cbd=ac, and hence, a c + b d = 0 a c + b d = 0 ac+bd=0a c+b d=0ac+bd=0. We conclude the dot product of perpendicular vectors is zero.
Both of these facts also follow from the geometric interpretation of the dot product shown in Figure 1.2. In this figure, we see that V 1 V 1 V_(1)V_{1}V1. V 2 V 2 V_(2)V_{2}V2 is the length of the projection of V 2 V 2 V_(2)V_{2}V2 onto V 1 V 1 V_(1)V_{1}V1, times the length of V 1 V 1 V_(1)V_{1}V1. Letting θ θ theta\thetaθ be the angle between these two vectors leads to an alternate way to compute dot products:
V 1 V 2 = | V 1 | | V 2 | cos θ V 1 V 2 = V 1 V 2 cos θ V_(1)*V_(2)=|V_(1)||V_(2)|cos thetaV_{1} \cdot V_{2}=\left|V_{1}\right|\left|V_{2}\right| \cos \thetaV1V2=|V1||V2|cosθ
To see this, note that the length L L LLL of the projection of V 2 V 2 V_(2)V_{2}V2 onto V 1 V 1 V_(1)V_{1}V1 is given by | V 2 | cos θ V 2 cos θ |V_(2)|cos theta\left|V_{2}\right| \cos \theta|V2|cosθ.
1.3. Suppose V 1 = a , b , V 2 = c , a V 1 = a , b , V 2 = c , a V_(1)=(:a,b:),V_(2)=(:c,a:)V_{1}=\langle a, b\rangle, V_{2}=\langle c, a\rangleV1=a,b,V2=c,a and θ θ theta\thetaθ is the angle between them. Show that
a c + b d = | V 1 | | V 2 | cos θ . a c + b d = V 1 V 2 cos θ . ac+bd=|V_(1)||V_(2)|cos theta.a c+b d=\left|V_{1}\right|\left|V_{2}\right| \cos \theta .ac+bd=|V1||V2|cosθ.
1.4. Use the dot product to compute the cosine of the angle between the vectors 1 , 2 1 , 2 (:1,2:)\langle 1,2\rangle1,2 and 4 , 2 4 , 2 (:4,2:)\langle 4,2\rangle4,2.
Another geometric quantity that we will need is the area of the parallelogram spanned by two vectors.
1.5. Suppose V 1 = a , b V 1 = a , b V_(1)=(:a,b:)V_{1}=\langle a, b\rangleV1=a,b and V 2 = c , a V 2 = c , a V_(2)=(:c,a:)V_{2}=\langle c, a\rangleV2=c,a. Show that the area of the parallelogram spanned by these vectors is | a d b c | | a d b c | |ad-bc||a d-b c||adbc|.
One common way to denote a set of vectors is by writing a matrix where the vectors appear as columns (or rows). The determinant of such a matrix is then defined to be the (signed) area of the parallelogram spanned by its column vectors. So, from the last exercise we have:
| a c b d | = a d b c . a      c b      d = a d b c . |[a,c],[b,d]|=ad-bc.\left|\begin{array}{ll} a & c \\ b & d \end{array}\right|=a d-b c .|acbd|=adbc.
Notice that this answer may be negative. This is because the determinant not only tells us area, but also something about the order of the vectors a , b a , b (:a,b:)\langle a, b\ranglea,b and c , a c , a (:c,a:)\langle c, a\ranglec,a.
Everything we have discussed above generalizes to higher dimensions. For example, if V 1 = a , b , c V 1 = a , b , c V_(1)=(:a,b,c:)V_{1}=\langle a, b, c\rangleV1=a,b,c, then the length of V 1 V 1 V_(1)V_{1}V1 is given by
| V 1 | = a 2 + b 2 + c 2 V 1 = a 2 + b 2 + c 2 |V_(1)|=sqrt(a^(2)+b^(2)+c^(2))\left|V_{1}\right|=\sqrt{a^{2}+b^{2}+c^{2}}|V1|=a2+b2+c2
If V 2 = d , e , f V 2 = d , e , f V_(2)=(:d,e,f:)V_{2}=\langle d, e, f\rangleV2=d,e,f, then
V 1 + V 2 = a + d , b + e , c + f . V 1 + V 2 = a + d , b + e , c + f . V_(1)+V_(2)=(:a+d,b+e,c+f:).V_{1}+V_{2}=\langle a+d, b+e, c+f\rangle .V1+V2=a+d,b+e,c+f.
The same geometric interpretation of addition (sliding V 2 V 2 V_(2)V_{2}V2 until its tail ends up at the tip of V 1 V 1 V_(1)V_{1}V1 ) holds in higher dimensions as well. The dot product also works as expected in higher dimensions:
V 1 V 2 = a d + b e + c f V 1 V 2 = a d + b e + c f V_(1)*V_(2)=ad+be+cfV_{1} \cdot V_{2}=a d+b e+c fV1V2=ad+be+cf
and its geometric interpretation as the projected length of V 2 V 2 V_(2)V_{2}V2, times the length of V 1 V 1 V_(1)V_{1}V1, holds.
It is a bit harder to show, but if V 3 = g , h , 1 V 3 = g , h , 1 V_(3)=(:g,h,1:)V_{3}=\langle g, h, 1\rangleV3=g,h,1, then the volume of the parallelpiped spanned by V 1 , V 2 V 1 , V 2 V_(1),V_(2)V_{1}, V_{2}V1,V2 and V 3 V 3 V_(3)V_{3}V3 is given by the absolute value of:
| a d g b e h c f i | = ( a e i + d h c + g b f ) ( a h f + d b i + g e c ) a      d      g b      e      h c      f      i = ( a e i + d h c + g b f ) ( a h f + d b i + g e c ) |[a,d,g],[b,e,h],[c,f,i]|=(aei+dhc+gbf)-(ahf+dbi+gec)\left|\begin{array}{lll} a & d & g \\ b & e & h \\ c & f & i \end{array}\right|=(a e i+d h c+g b f)-(a h f+d b i+g e c)|adgbehcfi|=(aei+dhc+gbf)(ahf+dbi+gec)
This is the formula for the determinant of a 3 × 3 3 × 3 3xx33 \times 33×3 matrix.
1.6. Find the volume of the parallelpiped spanned by the vectors 1 1 (:1\langle 11, 0 , 1 , 1 , 2 , 3 0 , 1 , 1 , 2 , 3 0,1:),(:1,2,3:)0,1\rangle,\langle 1,2,3\rangle0,1,1,2,3 and 2 , 5 , 3 2 , 5 , 3 (:2,5,3:)\langle 2,5,3\rangle2,5,3.

1.2 Functions of multiple variables

We denote by n n nnn the set of points with n n nnn-coordinates. If n n nnn is between 1 and 3 these spaces are very familiar. For example, 1 1 ^(1){ }^{1}1 is just the number line whose depiction hangs above every elementary school blackboard. The space 2 2 ^(2){ }^{2}2 is just the x y x y xyx yxy-plane that we employ so often in precalculus and calculus. And is, of course, familiar as the three-dimensional space that we feel like we experience every day (mathematicians and physicists debate whether or not this is really the space we live in).
The space R 4 R 4 R^(4)\mathbb{R}^{4}R4 is probably less familiar. One can think of the extra coordinate as time, or color, or anything else that gives more information. At some point we must just give up on visualization. There is no way to picture $ 20 $ 20 $^(20)\$^{20}$20. This does not mean it is useless. To model the stock market, for example, one may want to represent its state at a particular point in time as a point with as many coordinates as there are stocks.
Fortunately for us, if you really understand differential forms in dimensions up to three, then very little needs to be addressed to generalize to higher dimensions.
In this text we will often represent functions abstractly by saying how many numbers go into the function, and how many come out. So, if we write f : n m f : n m f:n rarr^(m)f: n \rightarrow{ }^{m}f:nm, we mean f f fff is a function whose input is a point with n n nnn coordinates and whose output is a point with m m mmm coordinates.
Some cases of this are familiar. For example, if y = f ( x ) y = f ( x ) y=f(x)y=f(x)y=f(x) is a typical function from Calculus I, then f : 1 1 f : 1 1 f:^(1)rarr1f:{ }^{1} \rightarrow 1f:11.
In this chapter, we focus on functions of the form f : R 2 f : R 2 f:R^(2)f: \mathbb{R}^{2}f:R2. These functions look something like z = f ( x , y ) z = f ( x , y ) z=f(x,y)z=f(x, y)z=f(x,y). To graph such a function we draw x x xxx-, y y yyy-, and z z zzz-axes in 3 3 ^(3){ }^{3}3 and plot all the points where the equation z = f ( x , y ) z = f ( x , y ) z=f(x,y)z=f(x, y)z=f(x,y) is true.
Computers can really help one visualize such graphs. It is worthwhile to play with any software package that will graph such functions. But it is equally worthwhile to learn a few techniques to sketch such graphs by hand.
The easiest way to begin to get a feel for a graph is by drawing its intersection with the coordinate planes. To sketch the intersection with the x z x z xzx zxz-plane, for example, set y y yyy equal to zero and graph the resulting function. Similarly, to sketch the intersection with the y z y z yzy zyz plane, set x x xxx equal to zero.
A similar approach involves sketching level curves. These are just the intersections of horizontal planes of the form z = n z = n z=nz=nz=n with the graph. To sketch such a curve, one simply plots the graph of n = f n = f n=fn=fn=f ( x , y x , y x,yx, yx,y ).
Putting all of this information together on one set of axes can be a challenge (see Figure 1.3). Some artistic ability and some ability to visualize three-dimensional shapes is helpful, but nothing substitutes for lots of practice.
1.7. Sketch the graphs of
1. z = 2 x 3 y . 2. z = x 2 + y 2 . 3. z = x y (compare with Figure 1.3). 4. z = x 2 + y 2 . 5. z = 1 x 2 + y 2 6. z = x 2 + y 2 + 1 . 7. z = x 2 + y 2 1 . 8 . z = cos ( x + y ) 9. z = cos ( x y ) 10. z = cos ( x 2 + y 2 ) . 11. z = e ( x 2 + y 2 ) .  1.  z = 2 x 3 y .  2.  z = x 2 + y 2 .  3.  z = x y  (compare with Figure 1.3).   4.  z = x 2 + y 2 .  5.  z = 1 x 2 + y 2  6.  z = x 2 + y 2 + 1 .  7.  z = x 2 + y 2 1 . 8 . z = cos ( x + y )  9.  z = cos ( x y )  10.  z = cos x 2 + y 2 .  11.  z = e x 2 + y 2 . {:[" 1. "z=2x-3y.],[" 2. "z=x^(2)+y^(2).],[" 3. "z=xy" (compare with Figure 1.3). "],[" 4. "z=sqrt(x^(2)+y^(2)).],[" 5. "z=(1)/(sqrt(x^(2)+y^(2)))],[" 6. "z=sqrt(x^(2)+y^(2)+1).],[" 7. "z=sqrt(x^(2)+y^(2)-1).],[8.z=cos(x+y)],[" 9. "z=cos(xy)],[" 10. "z=cos(x^(2)+y^(2)).],[" 11. "z=e^(-(x^(2)+y^(2))).]:}\begin{aligned} \text { 1. } z & =2 x-3 y . \\ \text { 2. } z & =x^{2}+y^{2} . \\ \text { 3. } z & =x y \text { (compare with Figure 1.3). } \\ \text { 4. } z & =\sqrt{x^{2}+y^{2}} . \\ \text { 5. } z & =\frac{1}{\sqrt{x^{2}+y^{2}}} \\ \text { 6. } z & =\sqrt{x^{2}+y^{2}+1} . \\ \text { 7. } z & =\sqrt{x^{2}+y^{2}-1} . \\ 8 . z & =\cos (x+y) \\ \text { 9. } z & =\cos (x y) \\ \text { 10. } z & =\cos \left(x^{2}+y^{2}\right) . \\ \text { 11. } z & =e^{-\left(x^{2}+y^{2}\right)} . \end{aligned} 1. z=2x3y. 2. z=x2+y2. 3. z=xy (compare with Figure 1.3).  4. z=x2+y2. 5. z=1x2+y2 6. z=x2+y2+1. 7. z=x2+y21.8.z=cos(x+y) 9. z=cos(xy) 10. z=cos(x2+y2). 11. z=e(x2+y2).

1.8. Find functions whose graphs are

  1. A plane through the origin at 45 : to both the x x xxx - and y y yyy-axes.
  2. The top half of a sphere of radius two.
  3. The top half of a torus centered around the z z zzz-axis (i.e., the tube of radius one, say, centered around a circle of radius two in the x y x y xyx yxy-plane).
  4. The top half of the cylinder of radius one which is centered around the line where the plane y = x y = x y=xy=xy=x meets the plane z = 0 z = 0 z=0z=0z=0.
You may find it helpful to check your answers to the above exercises with a computer graphing program.
Fig. 1.3. Several views of the graph of z = x 2 y 2 z = x 2 y 2 z=x^(2)-y^(2)z=x^{2}-y^{2}z=x2y2. The top two figures are the intersections with the x z x z xzx zxz - and y z y z yzy zyz-planes. The bottom left shows several level curves.



1.3 Multiple integrals

We now address the question of how to find the volume under the graph of a function f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) of two variables. Recall from Calculus I that we define the integral of a function g ( x ) g ( x ) g(x)g(x)g(x) of one variable on the interval [ 0 , a ] [ 0 , a ] [0,a][0, a][0,a] by the following steps:
  1. Choose a sequence of evenly spaced points { x i } i = 0 in n [ 0 , a ] x i i = 0  in  n [ 0 , a ] {x_(i)}_(i=0" in ")^(n)[0,a]\left\{x_{i}\right\}_{i=0 \text { in }}^{n}[0, a]{xi}i=0 in n[0,a] such that x 0 = 0 x 0 = 0 x_(0)=0x_{0}=0x0=0 and x n = a x n = a x_(n)=ax_{n}=axn=a.
  2. Let Δ x = x i + 1 x i Δ x = x i + 1 x i Delta x=x_(i+1)-x_(i)\Delta x=x_{i+1}-x_{i}Δx=xi+1xi.
  3. For each i compute g ( x i ) Δ x g x i Δ x g(x_(i))Delta xg\left(x_{i}\right) \Delta xg(xi)Δx.
  4. Sum over all i i iii.
  5. Take the limit as n n nnn goes to oo\infty.
The intuition is that each term in Step 3 above gives the area of a rectangle. Piecing all of the rectangles together gives an approximation for the function g ( x ) g ( x ) g(x)g(x)g(x), so the result of Step 4 is an approximation for the desired area. As n n nnn goes to oo\infty in Step 5, this approximation gets better and better.
Similar steps define the volume under f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y). Let R R RRR be the rectangle in the x y x y xyx yxy-plane with vertices at ( 0 , 0 ) , ( a , 0 ) , ( 0 , b ) ( 0 , 0 ) , ( a , 0 ) , ( 0 , b ) (0,0),(a,0),(0,b)(0,0),(a, 0),(0, b)(0,0),(a,0),(0,b) and ( a a aaa, b). We now perform the following steps:
  1. Choose sequences of evenly spaced points { x i } i = 0 and n { y j } j = 0 m x i i = 0  and  n y j j = 0 m {x_(i)}_(i=0" and ")^(n){y_(j)}_(j=0)^(m)\left\{x_{i}\right\}_{i=0 \text { and }}^{n}\left\{y_{j}\right\}_{j=0}^{m}{xi}i=0 and n{yj}j=0m such that x 0 = y 0 = 0 , x n = a x 0 = y 0 = 0 , x n = a x_(0)=y_(0)=0,x_(n)=ax_{0}=y_{0}=0, x_{n}=ax0=y0=0,xn=a and y m = b y m = b y_(m)=by_{m}=bym=b. This gives a lattice of points of the form ( x i , y j ) x i , y j (x_(i),y_(j))\left(x_{i}, y_{j}\right)(xi,yj) in R R RRR.
  2. Let Δ x = x i + 1 x i Δ x = x i + 1 x i Delta x=x_(i+1)-x_(i)\Delta x=x_{i+1}-x_{i}Δx=xi+1xi and Δ y = y j + 1 y j Δ y = y j + 1 y j Delta y=y_(j+1)-y_(j)\Delta y=y_{j+1}-y_{j}Δy=yj+1yj.
  3. For each i i iii and j j jjj compute f ( x i , y j ) Δ x Δ y f x i , y j Δ x Δ y f(x_(i),y_(j))Delta x Delta yf\left(x_{i}, y_{j}\right) \Delta x \Delta yf(xi,yj)ΔxΔy.
  4. Sum over all i i iii and j j jjj.
  5. Take the limit as n n nnn and m m mmm go to oo\infty.
These steps define f ( x , y ) d x d y f ( x , y ) d x d y f(x,y)dxdyf(x, y) d x d yf(x,y)dxdy. The intuition as to why this represents the desired volume is s s sss milar to that in the one variable case. In Step 3 we are computing the volume of a box whose base is a Δ x Δ x Delta x\Delta xΔx by Δ y Δ y Delta y\Delta yΔy rectangle, and whose height is f ( x i , y j ) f x i , y j f(x_(i),y_(j))f\left(x_{i}, y_{j}\right)f(xi,yj) (see Figure 1.4). Putting these boxes together approximates the function f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y), and this approximation gets better and better when n n nnn and m m mmm go to oo\infty.
It is important to understand the above definition from a theoretical point of view. Later in this text we will come back to it many times. Unfortunately, it is almost impossible to use this definition to compute any integrals. For this, we need an alternate point of view.
Instead of approximating f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) with boxes as above, we will now approximate it by "slabs" whose profiles look like slices by planes parallel to one of the coordinate planes (see Figure 1.6). To do this we carry out the following steps:
  1. Choose a sequence of evenly spaced points { x i } i = o s u c h t h a t n x i i = o s u c h t h a t n {x_(i)}_(i=osuchthat)^(n)\left\{x_{i}\right\}_{i=o s u c h ~ t h a t ~}^{n}{xi}i=osuch that n x 0 = 0 x 0 = 0 x_(0)=0x_{0}=0x0=0 and x n = a x n = a x_(n)=ax_{n}=axn=a.
  2. Let Δ x = x i + 1 x i Δ x = x i + 1 x i Delta x=x_(i+1)-x_(i)\Delta x=x_{i+1}-x_{i}Δx=xi+1xi.
  3. For each i compute
[ 0 b f ( x i , y ) d y ] Δ x . 0 b f x i , y d y Δ x . [int_(0)^(b)f(x_(i),y)dy]Delta x.\left[\int_{0}^{b} f\left(x_{i}, y\right) d y\right] \Delta x .[0bf(xi,y)dy]Δx.
  1. Sum over all i i iii.
  2. Take the limit as n n nnn goes to oo\infty. b b bbb
Note that in Step 3 the quantityo f ( x i j y ) d y f x i j y d y f(x_(ij)y)dyf\left(x_{i j} y\right) d yf(xijy)dy is exactly the area under the curve that you get when you slice the graph of f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) by
the plane parallel to the y z y z yzy zyz-plane at x = x i x = x i x=x_(i)x=x_{i}x=xi (see Figure 1.5). Multiplying by Δ x Δ x Delta x\Delta xΔx then gives the volume of a slab of thickness Δ x Δ x Delta x\Delta xΔx, with the same profile as this slice. Putting these slabs together still approximates the function f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y), and this approximation gets better and better as n n nnn goes to oo\infty (see Figure 1.6). The result is the following:
Fig. 1.4. Using boxes to approximate a function.
R f ( x , y ) d x d y = 0 a [ 0 b f ( x , y ) d y ] d x R f ( x , y ) d x d y = 0 a 0 b f ( x , y ) d y d x int_(R)f(x,y)dxdy=int_(0)^(a)[int_(0)^(b)f(x,y)dy]dx\int_{R} f(x, y) d x d y=\int_{0}^{a}\left[\int_{0}^{b} f(x, y) d y\right] d xRf(x,y)dxdy=0a[0bf(x,y)dy]dx
Of course, we could have added up volumes of the slabs that were parallel to the x z x z xzx zxz-plane instead. This process would have produced the following equality:
R f ( x , y ) d x d y = 0 b [ 0 a f ( x , y ) d x ] d y R f ( x , y ) d x d y = 0 b 0 a f ( x , y ) d x d y int_(R)f(x,y)dxdy=int_(0)^(b)[int_(0)^(a)f(x,y)dx]dy\int_{R} f(x, y) d x d y=\int_{0}^{b}\left[\int_{0}^{a} f(x, y) d x\right] d yRf(x,y)dxdy=0b[0af(x,y)dx]dy
Hence we see that Fubini's theorem must be true:
0 a 0 b f ( x , y ) d y d x = 0 b 0 a f ( x , y ) d x d y 0 a 0 b f ( x , y ) d y d x = 0 b 0 a f ( x , y ) d x d y int_(0)^(a)int_(0)^(b)f(x,y)dydx=int_(0)^(b)int_(0)^(a)f(x,y)dxdy\int_{0}^{a} \int_{0}^{b} f(x, y) d y d x=\int_{0}^{b} \int_{0}^{a} f(x, y) d x d y0a0bf(x,y)dydx=0b0af(x,y)dxdy
Fig. 1.5. The area A A AAA of the slice through x = x i x = x i x=x_(i)x=x_{i}x=xi is given byo f ( x i , f x i , f(x_(i,):}f\left(x_{i,}\right.f(xi, y) d y d y dyd ydy.
Fig. 1.6. Putting slabs together approximates the function f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y).
Note 1. Be aware that we have avoided very technical issues here such as continuity and convergence. For a rigorous treatment, see any standard text in multivariable calculus.
Example 1. To find the volume under the graph of f ( x , y ) = x y 2 f ( x , y ) = x y 2 f(x,y)=xy^(2)f(x, y)=x y^{2}f(x,y)=xy2 and above the rectangle R R RRR with vertices at ( 0 , 0 ) , ( 2 , 0 ) , ( 0 , 3 ) ( 0 , 0 ) , ( 2 , 0 ) , ( 0 , 3 ) (0,0),(2,0),(0,3)(0,0),(2,0),(0,3)(0,0),(2,0),(0,3) and ( 2 , 3 ) ( 2 , 3 ) (2,3)(2,3)(2,3) we compute:
R x y 2 d x d y = 0 3 0 2 x y 2 d x d y = 0 3 [ 1 2 x 2 y 2 | x = 0 2 ] d y = 0 3 2 y 2 d y = 18 R x y 2 d x d y = 0 3 0 2 x y 2 d x d y = 0 3 1 2 x 2 y 2 x = 0 2 d y = 0 3 2 y 2 d y = 18 {:[int_(R)xy^(2)dxdy=int_(0)^(3)int_(0)^(2)xy^(2)dxdy],[=int_(0)^(3)[(1)/(2)x^(2)y^(2)|_(x=0)^(2)]dy],[=int_(0)^(3)2y^(2)dy],[=18]:}\begin{aligned} \int_{R} x y^{2} d x d y & =\int_{0}^{3} \int_{0}^{2} x y^{2} d x d y \\ & =\int_{0}^{3}\left[\left.\frac{1}{2} x^{2} y^{2}\right|_{x=0} ^{2}\right] d y \\ & =\int_{0}^{3} 2 y^{2} d y \\ & =18 \end{aligned}Rxy2dxdy=0302xy2dxdy=03[12x2y2|x=02]dy=032y2dy=18
1.9. Let R R RRR be the rectangle in the x y x y xyx yxy-plane with vertices at ( 1,0 ), ( 2 , 0 ) , ( 1 , 3 ) ( 2 , 0 ) , ( 1 , 3 ) (2,0),(1,3)(2,0),(1,3)(2,0),(1,3) and ( 2 , 3 ) ( 2 , 3 ) (2,3)(2,3)(2,3). Integrate the following functions over R R RRR.
  1. x 2 y 2 x 2 y 2 x^(2)y^(2)x^{2} y^{2}x2y2
  2. 1 .
  3. x 2 + y 2 x 2 + y 2 x^(2)+y^(2)x^{2}+y^{2}x2+y2.
  4. x + 2 3 y x + 2 3 y sqrt(x+(2)/(3)y)\sqrt{x+\frac{2}{3} y}x+23y.

1.4 Partial derivatives

In this section, we begin to discuss tangent lines to the graph of a function of the form f : N 2 f : N 2 f:N^(2)f: \mathbb{N}^{2}f:N2. If we slice the graph of such a function with the plane parallel to the y z y z yzy zyz-plane, through the point ( x 0 , y 0 x 0 , y 0 x_(0),y_(0)x_{0}, y_{0}x0,y0 ), then we get a curve which represents some function of y y yyy. We can then ask, "What is the slope of the tangent line to this curve when y = y 0 y = y 0 y=y_(0)y=y_{0}y=y0 ?" The answer to this question is precisely the definition f o f y ( x 0 , y 0 ) f o f y x 0 , y 0 (del f)/(of del y)(x_(0),y_(0))\frac{\partial f}{o f \partial y}\left(x_{0}, y_{0}\right)fofy(x0,y0) (see Figure 1.7).
Example 2. Suppose f ( x , y ) = x y 2 f ( x , y ) = x y 2 f(x,y)=xy^(2)f(x, y)=x y^{2}f(x,y)=xy2. We wish to compute f y ( 2 , 3 ) f y ( 2 , 3 ) (del f)/(del y)(2,3)\frac{\partial f}{\partial y}(2,3)fy(2,3). The slice of the graph of f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y), parallel to the y z y z yzy zyz-plane, through the point ( 2,3 ), is given by substituting 2 for x x xxx. This gives us the function 2 y 2 2 y 2 2y^(2)2 y^{2}2y2. Differentiating with respect to y y yyy then gives 4 y 4 y 4y4 y4y. Plugging in 3 for y y yyy yields 12 .
If we instead wish to compute f y ( 4 , 3 ) f y ( 4 , 3 ) (del f)/(del y)(4,3)\frac{\partial f}{\partial y}(4,3)fy(4,3), we could go through the same steps. The slice through the point is the graph of 4 y 2 4 y 2 4y^(2)4 y^{2}4y2. Differentiating with respect to y y yyy gives 8 y 8 y 8y8 y8y. Evaulating at y = 3 y = 3 y=3y=3y=3 yields 24.
Fig. 1.7. The partial derivative with respect to y y yyy.
If we wish to repeat this many more times, it will be easier to leave the variable x x xxx in, but think of it as a constant. Hence, differentiating x y 2 x y 2 xy^(2)x y^{2}xy2 with respect to y y yyy gives 2 x y 2 x y 2xy2 x y2xy, and we can now plug in whatever numbers we want for x x xxx and y y yyy to obtain a final answer immediately.
Partial derivatives with respect to x x xxx are just as easy to compute. Geometrically, we think of this as giving the slope of a line tangent to the graph which is the slice parallel to the x z x z xzx zxz-plane. Algebraically, we think of y y yyy as a constant and take the derivative with respect to x x xxx. 1.10. Compute f x f x (del f)/(del x)\frac{\partial f}{\partial x}fx and f y f y (del f)/(del y)\frac{\partial f}{\partial y}fy.
  1. x 2 y 3 x 2 y 3 x^(2)y^(3)x^{2} y^{3}x2y3.
  2. sin ( x 2 y 3 ) sin x 2 y 3 sin(x^(2)y^(3))\sin \left(x^{2} y^{3}\right)sin(x2y3).
  3. x sin ( x y ) x sin ( x y ) x sin(xy)x \sin (x y)xsin(xy).
Notice that when you take a partial derivative you get another function of x x xxx and y y yyy. You can then do it again to find the second partials. These are denoted by: 1.11. Find all second partials for each of the functions in the previous exercise.
2 f x 2 = x ( f x ) 2 f x 2 = x f x (del^(2)f)/(delx^(2))=(del)/(del x)((del f)/(del x))\frac{\partial^{2} f}{\partial x^{2}}=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)2fx2=x(fx)
2 f y 2 = y ( f y ) 2 f x y = x ( f y ) 2 f y x = y ( f x ) 2 f y 2 = y f y 2 f x y = x f y 2 f y x = y f x {:[(del^(2)f)/(dely^(2))=(del)/(del y)((del f)/(del y))],[(del^(2)f)/(del x del y)=(del)/(del x)((del f)/(del y))],[(del^(2)f)/(del y del x)=(del)/(del y)((del f)/(del x))]:}\begin{aligned} \frac{\partial^{2} f}{\partial y^{2}} & =\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right) \\ \frac{\partial^{2} f}{\partial x \partial y} & =\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right) \\ \frac{\partial^{2} f}{\partial y \partial x} & =\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right) \end{aligned}2fy2=y(fy)2fxy=x(fy)2fyx=y(fx)
Note that amazingly, the "mixed" partials 2 f x y 2 f x y (del^(2)f)/(del x del y)\frac{\partial^{2} f}{\partial x \partial y}2fxy and 2 f y x 2 f y x (del^(2)f)/(del y del x)\frac{\partial^{2} f}{\partial y \partial x}2fyx are always equal. This is not a coincidence! Somehow the mixed partials measure the "twisting" of the graph, and this is the same from every direction.

1.5 Gradients

Let's look back to Figure 1.7. What if we sliced the graph of f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) with some vertical plane through the point ( x 0 , y 0 x 0 , y 0 x_(0),y_(0)x_{0}, y_{0}x0,y0 ) that was not parallel to the x z x z xzx zxz - or y z y z yzy zyz-planes, as in Figure 1.8? How could we compute the slope then?
Fig. 1.8. A directional derivative.

( x 0 , y 0 ) x 0 , y 0 (x_(0),y_(0))\left(x_{0}, y_{0}\right)(x0,y0)
To answer this, visualize the set of all lines tangent to the graph of f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) at the point ( x 0 , y 0 x 0 , y 0 x_(0),y_(0)x_{0}, y_{0}x0,y0 ). The is a tangent plane.
The equation for a plane through the origin in 3 3 ^(3){ }^{3}3 is of the form z z zzz = m x x + m y y = m x x + m y y =m_(x)x+m_(y)y=m_{x} x+m_{y} y=mxx+myy. Notice that the intersection of such a plane with the x z x z xz-x z-xz
plane is the graph of z = m x x z = m x x z=m_(x)xz=m_{x} xz=mxx. Hence, m x m x m_(x)m_{x}mx is the slope of this line of intersection. Similarly, the quantity m y m y m_(y)m_{y}my is the slope of the line which is the intersection with the y z y z yzy zyz-plane.
To get a plane through the point ( x 0 , y 0 , f ( x 0 , y 0 ) x 0 , y 0 , f x 0 , y 0 x_(0),y_(0),f(x_(0),y_(0))x_{0}, y_{0}, f\left(x_{0}, y_{0}\right)x0,y0,f(x0,y0) ), we can translate the origin to this point by replacing x x xxx with x x 0 , y x x 0 , y x-x_(0),yx-x_{0}, yxx0,y with y y yyy y 0 y 0 y_(0)y_{0}y0 and z z zzz with z f ( x 0 , y 0 ) z f x 0 , y 0 z-f(x_(0),y_(0))z-f\left(x_{0}, y_{0}\right)zf(x0,y0) :
z f ( x 0 , y 0 ) = m x ( x x 0 ) + m y ( y y 0 ) . z f x 0 , y 0 = m x x x 0 + m y y y 0 . z-f(x_(0),y_(0))=m_(x)(x-x_(0))+m_(y)(y-y_(0)).z-f\left(x_{0}, y_{0}\right)=m_{x}\left(x-x_{0}\right)+m_{y}\left(y-y_{0}\right) .zf(x0,y0)=mx(xx0)+my(yy0).
Since we want this to actually be a tangent plane, it follows that m x m x m_(x)m_{x}mx must be equal to f x f x (del f)/(del x)\frac{\partial f}{\partial x}fx and m y m y mym ymy must be f y f y (del f)/(del y)\frac{\partial f}{\partial y}fy. Hence, the equation of the tangent plane T T TTT is given by where f x f y f x f y (del f)/(del x)(del f)/(del y)\frac{\partial f}{\partial x} \frac{\partial f}{\partial y}fxfy, and f f fff are all evaluated at the point ( x 0 , y 0 x 0 , y 0 x_(0),y_(0)x_{0}, y_{0}x0,y0 ).
T ( x , y ) = f x ( x x 0 ) + f y ( y y 0 ) + f T ( x , y ) = f x x x 0 + f y y y 0 + f T(x,y)=(del f)/(del x)(x-x_(0))+(del f)/(del y)(y-y_(0))+fT(x, y)=\frac{\partial f}{\partial x}\left(x-x_{0}\right)+\frac{\partial f}{\partial y}\left(y-y_{0}\right)+fT(x,y)=fx(xx0)+fy(yy0)+f
Now, suppose P P PPP is the vertical plane through the point ( x 0 , y 0 x 0 , y 0 x_(0),y_(0)x_{0}, y_{0}x0,y0 ) depicted in Figure 1.9. Let / / //// denote the line where P P PPP intersects the x y x y xyx yxy-plane. The tangent line L L LLL to the graph of f f fff, which lies above I I III, is also the line contained in T T TTT, which lies above I I III. To figure out the slope of L L LLL we will simply compute "rise over run."
Suppose / / //// contains the vector V = a , b V = a , b V=(:a,b:)V=\langle a, b\rangleV=a,b, where V = 1 V = 1 ∣V=1\mid V=1V=1. Then two points on I I III, a distance of 1 apart, are ( x 0 , y 0 x 0 , y 0 x_(0),y_(0)x_{0}, y_{0}x0,y0 ) and ( x 0 + a , y 0 + b x 0 + a , y 0 + b x_(0)+a,y_(0)+bx_{0}+a, y_{0}+bx0+a,y0+b ). Thus the "run" will be equal to 1 . The "rise" is the difference between T ( x 0 , y 0 ) T x 0 , y 0 T(x_(0),y_(0))T\left(x_{0}, y_{0}\right)T(x0,y0) and T ( x 0 + a , y 0 + b ) T x 0 + a , y 0 + b T(x_(0)+a,y_(0)+b)T\left(x_{0}+a, y_{0}+b\right)T(x0+a,y0+b), which we compute as follows:
T ( x 0 + a , y 0 + b ) T ( x 0 , y 0 ) = [ f x ( x 0 + a x 0 ) + f y ( y 0 + b y 0 ) + f ] [ f x ( x 0 x 0 ) + f y ( y 0 y 0 ) + f ] = a f x + b f y . T x 0 + a , y 0 + b T x 0 , y 0 = f x x 0 + a x 0 + f y y 0 + b y 0 + f f x x 0 x 0 + f y y 0 y 0 + f = a f x + b f y . {:[T(x_(0)+a,y_(0)+b)-T(x_(0),y_(0))],[=[(del f)/(del x)(x_(0)+a-x_(0))+(del f)/(del y)(y_(0)+b-y_(0))+f]],[-[(del f)/(del x)(x_(0)-x_(0))+(del f)/(del y)(y_(0)-y_(0))+f]],[=a(del f)/(del x)+b(del f)/(del y).]:}\begin{aligned} T\left(x_{0}+a, y_{0}+b\right) & -T\left(x_{0}, y_{0}\right) \\ = & {\left[\frac{\partial f}{\partial x}\left(x_{0}+a-x_{0}\right)+\frac{\partial f}{\partial y}\left(y_{0}+b-y_{0}\right)+f\right] } \\ & -\left[\frac{\partial f}{\partial x}\left(x_{0}-x_{0}\right)+\frac{\partial f}{\partial y}\left(y_{0}-y_{0}\right)+f\right] \\ = & a \frac{\partial f}{\partial x}+b \frac{\partial f}{\partial y} . \end{aligned}T(x0+a,y0+b)T(x0,y0)=[fx(x0+ax0)+fy(y0+by0)+f][fx(x0x0)+fy(y0y0)+f]=afx+bfy.
Since the slope of L L LLL is "rise" over "run," and the "run" equals 1 , we conclude the slope of L L LLL is a f x + b f y a f x + b f y ^(a(del f)/(del x))+b(del f)/(del y){ }^{a \frac{\partial f}{\partial x}}+b \frac{\partial f}{\partial y}afx+bfy, where f x f x (del f)/(del x)\frac{\partial f}{\partial x}fx and f y f y (del f)/(del y)\frac{\partial f}{\partial y}fy are evaluated at the point ( x 0 , y 0 x 0 , y 0 x_(0),y_(0)x_{0}, y_{0}x0,y0 ).
1.12. Suppose f ( x , y ) = x 2 y 3 f ( x , y ) = x 2 y 3 f(x,y)=x^(2)y^(3)f(x, y)=x^{2} y^{3}f(x,y)=x2y3. Compute the slope of the line tangent to f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y), at the point ( 2 , 1 ) ( 2 , 1 ) (2,1)(2,1)(2,1), in the direction ( 2 2 , 2 2 ) 2 2 , 2 2 ((sqrt2)/(2),-(sqrt2)/(2))\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)(22,22).
1.13. Let f ( x , y ) = x y + x 2 y + 4 f ( x , y ) = x y + x 2 y + 4 f(x,y)=xy+x-2y+4f(x, y)=x y+x-2 y+4f(x,y)=xy+x2y+4. Find the slope of the tangent line to the graph of f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y), in the direction of 1 , 2 1 , 2 (:1,2:)\langle 1,2\rangle1,2, at the point ( 0 , 1).
Fig. 1.9. Computing the slope of the tangent line L L LLL.
The quantity a f x ( x 0 , y 0 ) + b f y ( x 0 , y 0 ) is a f x x 0 , y 0 + b f y x 0 , y 0 is  ^(a)(del f)/(del x)(x_(0),y_(0))+b(del f)/(del y)(x_(0),y_(0))_("is "){ }^{a} \frac{\partial f}{\partial x}\left(x_{0}, y_{0}\right)+b \frac{\partial f}{\partial y}\left(x_{0}, y_{0}\right)_{\text {is }}afx(x0,y0)+bfy(x0,y0)is  defined to be the directional derivative of f f fff, at the point ( x 0 , y 0 ) x 0 , y 0 (x_(0),y_(0))\left(x_{0}, y_{0}\right)(x0,y0), in the direction V V VVV. We will adopt the notation V f ( x 0 , y 0 ) V f x 0 , y 0 grad_(V)f(x_(0),y_(0))\nabla_{V} f\left(x_{0}, y_{0}\right)Vf(x0,y0) for this quantity.
Let f ( x , y ) = x y 2 f ( x , y ) = x y 2 f(x,y)=xy^(2)f(x, y)=x y^{2}f(x,y)=xy2. Let's compute the directional derivative of f f fff, at the point ( 2 , 3 ) ( 2 , 3 ) (2,3)(2,3)(2,3), in the direction V = 1 , 5 V = 1 , 5 V=(:1,5:)V=\langle 1,5\rangleV=1,5. We compute:
V f ( 2 , 3 ) = 1 f x ( 2 , 3 ) + 5 f y ( 2 , 3 ) = 1 3 2 + 5 2 2 3 = 69 . V f ( 2 , 3 ) = 1 f x ( 2 , 3 ) + 5 f y ( 2 , 3 ) = 1 3 2 + 5 2 2 3 = 69 . {:[grad_(V)f(2","3)=1(del f)/(del x)(2","3)+5(del f)/(del y)(2","3)],[=1*3^(2)+5*2*2*3],[=69.]:}\begin{aligned} \nabla_{V} f(2,3) & =1 \frac{\partial f}{\partial x}(2,3)+5 \frac{\partial f}{\partial y}(2,3) \\ & =1 \cdot 3^{2}+5 \cdot 2 \cdot 2 \cdot 3 \\ & =69 . \end{aligned}Vf(2,3)=1fx(2,3)+5fy(2,3)=132+5223=69.
Is 69 the slope of the tangent line to some curve that we get when we intersect the graph of x y 2 x y 2 xy^(2)x y^{2}xy2 with some plane? What this number represents is the rate of change of f f fff, as we walk along the line / of Figure 1.9, with speed И И ∣И\mid ИИ. To find the desired slope we would have to walk with speed one. Hence, the directional derivative only
represents a slope when V = 1 V = 1 ∣V=1\mid V=1V=1. Let's at least see if this agrees with what we previously found.
If we stand at the point ( x 0 , y 0 x 0 , y 0 x_(0),y_(0)x_{0}, y_{0}x0,y0 ), walk in the direction 1 , 0 1 , 0 (:1,0:)\langle 1,0\rangle1,0 and ask what the rate of change of f f fff is, we obtain the following answer:
( 1 , 0 ) f ( x 0 , y 0 ) = 1 f x ( x 0 , y 0 ) + 0 f y ( x 0 , y 0 ) = f x ( x 0 , y 0 ) . ( 1 , 0 ) f x 0 , y 0 = 1 f x x 0 , y 0 + 0 f y x 0 , y 0 = f x x 0 , y 0 . grad_((1,0))f(x_(0),y_(0))=1(del f)/(del x)(x_(0),y_(0))+0(del f)/(del y)(x_(0),y_(0))=(del f)/(del x)(x_(0),y_(0)).\nabla_{(1,0)} f\left(x_{0}, y_{0}\right)=1 \frac{\partial f}{\partial x}\left(x_{0}, y_{0}\right)+0 \frac{\partial f}{\partial y}\left(x_{0}, y_{0}\right)=\frac{\partial f}{\partial x}\left(x_{0}, y_{0}\right) .(1,0)f(x0,y0)=1fx(x0,y0)+0fy(x0,y0)=fx(x0,y0).
This certainly agrees with our interpretation of f f (del f)/()\frac{\partial f}{}f as a slope. If we repeat this with the vector 2 , 0 2 , 0 (:2,0:)\langle 2,0\rangle2,0, then we find out how fast f f fff changes when we walk twice as fast in the same direction:
( 2 , 0 ) f ( x 0 , y 0 ) = 2 f x ( x 0 , y 0 ) + 0 f y ( x 0 , y 0 ) = 2 f x ( x 0 , y 0 ) . ( 2 , 0 ) f x 0 , y 0 = 2 f x x 0 , y 0 + 0 f y x 0 , y 0 = 2 f x x 0 , y 0 . grad_((2,0))f(x_(0),y_(0))=2(del f)/(del x)(x_(0),y_(0))+0(del f)/(del y)(x_(0),y_(0))=2(del f)/(del x)(x_(0),y_(0)).\nabla_{(2,0)} f\left(x_{0}, y_{0}\right)=2 \frac{\partial f}{\partial x}\left(x_{0}, y_{0}\right)+0 \frac{\partial f}{\partial y}\left(x_{0}, y_{0}\right)=2 \frac{\partial f}{\partial x}\left(x_{0}, y_{0}\right) .(2,0)f(x0,y0)=2fx(x0,y0)+0fy(x0,y0)=2fx(x0,y0).
As expected, f f fff now changes twice as fast.
To proceed further, we write the definition of V f V f grad_(V)f\nabla_{V} fVf as a dot product:
( a , b ) f ( x 0 , y 0 ) = a f x + b f y = f x , f y a , b . ( a , b ) f x 0 , y 0 = a f x + b f y = f x , f y a , b . grad_((a,b))f(x_(0),y_(0))=a(del f)/(del x)+b(del f)/(del y)=(:(del f)/(del x),(del f)/(del y):)*(:a,b:).\nabla_{(a, b)} f\left(x_{0}, y_{0}\right)=a \frac{\partial f}{\partial x}+b \frac{\partial f}{\partial y}=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right\rangle \cdot\langle a, b\rangle .(a,b)f(x0,y0)=afx+bfy=fx,fya,b.
Using this notation we obtain the following formula:
V f ( x 0 , y 0 ) = f ( x 0 , y 0 ) V V f x 0 , y 0 = f x 0 , y 0 V grad_(V)f(x_(0),y_(0))=grad f(x_(0),y_(0))*V\nabla_{V} f\left(x_{0}, y_{0}\right)=\nabla f\left(x_{0}, y_{0}\right) \cdot VVf(x0,y0)=f(x0,y0)V
Note that this dot product is greatest when V V VVV points in the same direction as f f grad f\nabla ff. This fact leads us to the geometric significance of the gradient vector. Think of f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) as a function which represents the altitude in some mountain range, given a location in longitude x x xxx
and latitude y y yyy. Now, if all you know is f f fff and your location x x xxx and y y yyy, and you want to figure out which way "uphill" is, all you have to do is point yourself in the direction of f f grad f\nabla ff.
What if you wanted to know what the slope was in the direction of steepest ascent? You would have to compute the directional derivative, using a vector of length one which points in the same direction as f f grad f\nabla ff. Such a vector is easy to find: U = f | f | U = f | f | U=(grad f)/(|grad f|)U=\frac{\nabla f}{|\nabla f|}U=f|f|. Now we compute this slope:
U f = f U = f f | f | = 1 | f | ( f f ) = 1 | f | | f | 2 = | f | . U f = f U = f f | f | = 1 | f | ( f f ) = 1 | f | | f | 2 = | f | . {:[grad_(U)f=grad f*U],[=grad f*(grad f)/(|grad f|)],[=(1)/(|grad f|)(grad f*grad f)],[=(1)/(|grad f|)|grad f|^(2)],[=|grad f|.]:}\begin{aligned} \nabla_{U} f & =\nabla f \cdot U \\ & =\nabla f \cdot \frac{\nabla f}{|\nabla f|} \\ & =\frac{1}{|\nabla f|}(\nabla f \cdot \nabla f) \\ & =\frac{1}{|\nabla f|}|\nabla f|^{2} \\ & =|\nabla f| . \end{aligned}Uf=fU=ff|f|=1|f|(ff)=1|f||f|2=|f|.
Hence, the magnitude of the gradient vector represents the largest slope of a tangent line through a particular point.
1.14. Let f ( x , y ) = x y 2 f ( x , y ) = x y 2 f(x,y)=xy^(2)f(x, y)=x y^{2}f(x,y)=xy2.
  1. Compute f f grad f\nabla ff.
  2. Use your answer to the previous question to compute 1 , 5 1 , 5 grad(:_(1,5):)\nabla\left\langle_{1,5}\right\rangle1,5 f ( 2 , 3 ) f ( 2 , 3 ) f(2,3)f(2,3)f(2,3).
  3. Find a vector of length one that points in the direction of steepest ascent, at the point ( 2 , 3 ) ( 2 , 3 ) (2,3)(2,3)(2,3).
  4. What is the largest slope of a tangent line to the graph of f f fff when ( x , y ) = ( 2 , 3 ) ( x , y ) = ( 2 , 3 ) (x,y)=(2,3)(x, y)=(2,3)(x,y)=(2,3) ?
    1.15. Suppose ( x 0 , y 0 x 0 , y 0 x_(0),y_(0)x_{0}, y_{0}x0,y0 ) is a point where f f grad f\nabla ff is non-zero and let n = n = n=n=n= f ( x 0 , y 0 ) f x 0 , y 0 f(x_(0),y_(0))f\left(x_{0}, y_{0}\right)f(x0,y0). Show that the vector f ( x 0 , y 0 ) f x 0 , y 0 grad f(x_(0),y_(0))\nabla f\left(x_{0}, y_{0}\right)f(x0,y0) is perpendicular to the set of points ( x , y ) ( x , y ) (x,y)(x, y)(x,y) such that f ( x , y ) = n f ( x , y ) = n f(x,y)=nf(x, y)=nf(x,y)=n (i.e., a level curve).
    1.16. For each of the following functions f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) :
  • Compute f ( 0 , 0 ) f ( 0 , 0 ) grad f(0,0)\nabla f(0,0)f(0,0).
  • What does this answer tell you about the slope of the lines tangent to the graph of f f fff at ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0) ?
  • Compute all second partials at ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0).
  • At the point ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0) compute
D ( x , y ) = | 2 f x 2 f 2 f y x 2 f y 2 y | = 2 f x 2 2 f y 2 2 f x y 2 f y x . D ( x , y ) = 2 f x 2 f 2 f y x 2 f y 2 y = 2 f x 2 2 f y 2 2 f x y 2 f y x . D(x,y)=|[(del^(2)f)/(delx^(2))f],[(del^(2)f)/(del y del x)(del^(2)f)/(dely^(2)y)]|=(del^(2)f)/(delx^(2))(del^(2)f)/(dely^(2))-(del^(2)f)/(del x del y)(del^(2)f)/(del y del x).D(x, y)=\left|\begin{array}{l} \frac{\partial^{2} f}{\partial x^{2}} f \\ \frac{\partial^{2} f}{\partial y \partial x} \frac{\partial^{2} f}{\partial y^{2} y} \end{array}\right|=\frac{\partial^{2} f}{\partial x^{2}} \frac{\partial^{2} f}{\partial y^{2}}-\frac{\partial^{2} f}{\partial x \partial y} \frac{\partial^{2} f}{\partial y \partial x} .D(x,y)=|2fx2f2fyx2fy2y|=2fx22fy22fxy2fyx.
  • Describe the shape of the graph of f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) near ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0).
1. x 2 + y 2 2. x 2 y 2 3. x 2 y 2 4. x y  1.  x 2 + y 2  2.  x 2 y 2  3.  x 2 y 2  4.  x y {:[" 1. "x^(2)+y^(2)],[" 2. "-x^(2)-y^(2)],[" 3. "x^(2)-y^(2)],[" 4. "xy]:}\begin{aligned} & \text { 1. } x^{2}+y^{2} \\ & \text { 2. }-x^{2}-y^{2} \\ & \text { 3. } x^{2}-y^{2} \\ & \text { 4. } x y \end{aligned} 1. x2+y2 2. x2y2 3. x2y2 4. xy
1.17. A function f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) is said to have a critical point at ( x 0 , y 0 ) x 0 , y 0 (x_(0),y_(0))\left(x_{0}, y_{0}\right)(x0,y0) if grad\nabla f ( x 0 , y 0 ) = 0 , 0 f x 0 , y 0 = 0 , 0 f(x_(0),y_(0))=(:0,0:)f\left(x_{0}, y_{0}\right)=\langle 0,0\ranglef(x0,y0)=0,0. Based on the previous problem, hypothesize about whether the graph of z = f ( x , y ) z = f ( x , y ) z=f(x,y)z=f(x, y)z=f(x,y) has a maximum, minimum, or saddle at ( x 0 , y 0 ) x 0 , y 0 (x_(0),y_(0))\left(x_{0}, y_{0}\right)(x0,y0) if f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) has a critical point at ( x 0 , y 0 ) x 0 , y 0 (x_(0),y_(0))\left(x_{0}, y_{0}\right)(x0,y0),
  1. D ( x 0 , y 0 ) > 0 D x 0 , y 0 > 0 D(x_(0),y_(0)) > 0D\left(x_{0}, y_{0}\right)>0D(x0,y0)>0 and 2 f x 2 < 0 2 f x 2 < 0 (del^(2)f)/(delx^(2)) < 0\frac{\partial^{2} f}{\partial x^{2}}<02fx2<0.
  2. D ( x 0 , y 0 ) > 0 D x 0 , y 0 > 0 D(x_(0),y_(0)) > 0D\left(x_{0}, y_{0}\right)>0D(x0,y0)>0 and 2 f x 2 > 0 2 f x 2 > 0 (del^(2)f)/(delx^(2)) > 0\frac{\partial^{2} f}{\partial x^{2}}>02fx2>0.
  3. D ( x 0 , y 0 ) < 0 D x 0 , y 0 < 0 D(x_(0),y_(0)) < 0D\left(x_{0}, y_{0}\right)<0D(x0,y0)<0.
    1.18. Find functions f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) such that D ( 0 , 0 ) = 0 D ( 0 , 0 ) = 0 D(0,0)=0D(0,0)=0D(0,0)=0 and at ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0) the graph of z = f ( x , y ) z = f ( x , y ) z=f(x,y)z=f(x, y)z=f(x,y) has a
  1. Minimum.
  2. Maximum.
  3. Saddle.

2

Parameterizations

2.1 Parameterized curves in 2

Given a curve C C CCC in a parameterization for C C CCC is a (one-to-one, onto, differentiable) function of the form φ : 1 C φ : 1 C varphi:^(1)rarr C\varphi:{ }^{1} \rightarrow Cφ:1C.
Example 3. The function φ ( t ) = ( cos t φ ( t ) = ( cos t varphi(t)=(cos t\varphi(t)=(\cos tφ(t)=(cost, sin t ) sin t ) sin t)\sin t)sint), where 0 t < 2 π 0 t < 2 π 0 <= t < 2pi0 \leq t<2 \pi0t<2π, is a parameterization for the circle of radius 1 . Another parameterization for the same circle is Ψ ( t ) = ( cos 2 t Ψ ( t ) = ( cos 2 t Psi(t)=(cos 2t\Psi(t)=(\cos 2 tΨ(t)=(cos2t, sin 2 t ) sin 2 t ) sin 2t)\sin 2 t)sin2t), where 0 t < π 0 t < π 0 <= t < pi0 \leq t<\pi0t<π. The difference between these two parameterizations is that as t t ttt increases, the image of Ψ ( t ) Ψ ( t ) Psi(t)\Psi(t)Ψ(t) moves twice as fast around the circle as the image of φ ( t ) φ ( t ) varphi(t)\varphi(t)φ(t).
2.1. A function of the form φ ( t ) = ( a t + c , b t + d ) φ ( t ) = ( a t + c , b t + d ) varphi(t)=(at+c,bt+d)\varphi(t)=(a t+c, b t+d)φ(t)=(at+c,bt+d) is a parameterization of a line.
  1. What is the slope of the line parameterized by φ φ varphi\varphiφ ?
  2. How does this line compare to the one parameterized by Ψ ( t ) Ψ ( t ) Psi(t)\Psi(t)Ψ(t) = ( a t , b t ) = ( a t , b t ) =(at,bt)=(a t, b t)=(at,bt) ?
    2.2. Draw the curves given by the following parameterizations:
  3. ( t , t 2 ) t , t 2 (t,t^(2))\left(t, t^{2}\right)(t,t2), where 0 t 1 0 t 1 0 <= t <= 10 \leq t \leq 10t1.
  4. ( t 2 , t 3 ) t 2 , t 3 (t^(2),t^(3))\left(t^{2}, t^{3}\right)(t2,t3), where 0 t 1 0 t 1 0 <= t <= 10 \leq t \leq 10t1.
  5. ( 2 cos t , 3 sin t ) ( 2 cos t , 3 sin t ) (2cos t,3sin t)(2 \cos t, 3 \sin t)(2cost,3sint), where 0 t 2 π 0 t 2 π 0 <= t <= 2pi0 \leq t \leq 2 \pi0t2π.
  6. ( cos 2 t , sin 3 t ) ( cos 2 t , sin 3 t ) (cos 2t,sin 3t)(\cos 2 t, \sin 3 t)(cos2t,sin3t), where 0 t 2 π 0 t 2 π 0 <= t <= 2pi0 \leq t \leq 2 \pi0t2π.
  7. ( t cos t , t sin t ) ( t cos t , t sin t ) (t cos t,t sin t)(t \cos t, t \sin t)(tcost,tsint), where 0 t 2 π 0 t 2 π 0 <= t <= 2pi0 \leq t \leq 2 \pi0t2π.
Given a curve, it can be very difficult to find a parameterization. There are many ways of approaching the problem, but nothing which always works. Here are a few hints:
  1. If C C CCC is the graph of a function y = f ( x ) y = f ( x ) y=f(x)y=f(x)y=f(x), then φ ( t ) = ( t , f ( t ) ) φ ( t ) = ( t , f ( t ) ) varphi(t)=(t,f(t))\varphi(t)=(t, f(t))φ(t)=(t,f(t)) is a parameterization of C C CCC. Notice that the y y yyy-coordinate of every point in the image of this parameterization is obtained from the x x xxx-coordinate by applying the function f f fff.
  2. If one has a polar equation for a curve like r = f ( θ ) r = f ( θ ) r=f(theta)r=f(\theta)r=f(θ), then, since x = r cos θ x = r cos θ x=r cos thetax=r \cos \thetax=rcosθ and y = r sin θ y = r sin θ y=r sin thetay=r \sin \thetay=rsinθ, we get a parameterization of the form φ ( θ ) = ( f ( θ ) cos θ , f ( θ ) sin θ ) φ ( θ ) = ( f ( θ ) cos θ , f ( θ ) sin θ ) varphi(theta)=(f(theta)cos theta,f(theta)sin theta)\varphi(\theta)=(f(\theta) \cos \theta, f(\theta) \sin \theta)φ(θ)=(f(θ)cosθ,f(θ)sinθ).
Example 4. The top half of a circle of rad us one is the graph of y = y = y=y=y= 1 x 2 1 x 2 sqrt()1-x^(2)\sqrt{ } 1-x^{2}1x2. Hence a parameterization for this is ( t , 1 t 2 ) t , 1 t 2 (t,sqrt()1-t^(2))\left(t, \sqrt{ } 1-t^{2}\right)(t,1t2), where 1 t 1 t -1 <= t <=-1 \leq t \leq1t 1. This figure is also the graph of the polar equation r = 1 , 0 θ π r = 1 , 0 θ π r=1,0 <= theta <= pir=1,0 \leq \theta \leq \pir=1,0θπ, hence the parameterization ( cos t , sin t cos t , sin t cos t,sin t\cos t, \sin tcost,sint ), where 0 t π 0 t π 0 <= t <= pi0 \leq t \leq \pi0tπ.
2.3. Sketch and find parameterizations for the curves described by:
  1. The graph of the polar equation r = cos θ r = cos θ r=cos thetar=\cos \thetar=cosθ.
  2. The graph of y = sin x y = sin x y=sin xy=\sin xy=sinx.
    2.4. Find a parameterization for the line segment which connects the point ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1) to the point ( 2 , 5 ) ( 2 , 5 ) (2,5)(2,5)(2,5).
Parameterized curves may be familiar from a second semester calculus class. Often in these classes one learns how to calculate the slope of a tangent line to the curve. But usually one does not discuss the derivative of the parameterization itself. One reason is that the derivative is actually a vector. If φ ( t ) = ( f ( t ) , g ( t ) ) φ ( t ) = ( f ( t ) , g ( t ) ) varphi(t)=(f(t),g(t))\varphi(t)=(f(t), g(t))φ(t)=(f(t),g(t)), then
ϕ ( t ) = d ϕ d t = d d t ( f ( t ) , g ( t ) ) = f ( t ) , g ( t ) . ϕ ( t ) = d ϕ d t = d d t ( f ( t ) , g ( t ) ) = f ( t ) , g ( t ) . phi^(')(t)=(d phi)/(dt)=(d)/(dt)(f(t),g(t))=(:f^(')(t),g^(')(t):).\phi^{\prime}(t)=\frac{d \phi}{d t}=\frac{d}{d t}(f(t), g(t))=\left\langle f^{\prime}(t), g^{\prime}(t)\right\rangle .ϕ(t)=dϕdt=ddt(f(t),g(t))=f(t),g(t).
This vector has important geometric significance. The slope of a line containing this vector when t = t 0 t = t 0 t=t_(0)t=t_{0}t=t0 is the same as the slope of the line tangent to the curve at the point φ ( t 0 ) φ t 0 varphi(t_(0))\varphi\left(t_{0}\right)φ(t0). The magnitude (length) of this vector gives one a concept of the speed of the point
φ ( t ) φ ( t ) varphi(t)\varphi(t)φ(t) as t t ttt is increases through t 0 t 0 t_(0)t_{0}t0. For convenience, one often draws the vector φ ( t 0 ) φ t 0 varphi^(')(t_(0))\varphi^{\prime}\left(t_{0}\right)φ(t0) based at the point φ ( t 0 ) φ t 0 varphi_((t_(0)))\varphi_{\left(t_{0}\right)}φ(t0) (see Figure 2.1).
2.5. Let φ ( t ) = ( cos t , sin t ) φ ( t ) = ( cos t , sin t ) varphi(t)=(cos t,sin t)\varphi(t)=(\cos t, \sin t)φ(t)=(cost,sint) (where 0 t π ) 0 t π ) 0 <= t <= pi)0 \leq t \leq \pi)0tπ) and Ψ ( t ) = ( t , 1 t 2 ) Ψ ( t ) = t , 1 t 2 Psi(t)=(t,sqrt()1-t^(2))\Psi(t)=\left(t, \sqrt{ } 1-t^{2}\right)Ψ(t)=(t,1t2) (where 1 t 1 1 t 1 -1 <= t <= 1-1 \leq t \leq 11t1 ) be parameterizations of the top half of the unit circle. Sketch the vectors d ϕ d t d ϕ d t (d phi)/(dt)\frac{d \phi}{d t}dϕdt and d ψ d t d ψ d t (d psi)/(dt)\frac{d \psi}{d t}dψdt at the points ( 2 2 , 2 2 ) , ( 0 , 1 ) 2 2 , 2 2 , ( 0 , 1 ) ((sqrt2)/(2),(sqrt2)/(2)),(0,1)\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right),(0,1)(22,22),(0,1) and ( 2 2 , 2 2 ) 2 2 , 2 2 (-(sqrt2)/(2),(sqrt2)/(2))\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)(22,22).
2.6. Let C C CCC be the set of points in 2 2 ^(2){ }^{2}2 that satisfies the equation x = x = x=x=x= y 2 y 2 y^(2)y^{2}y2.
  1. Find a parameterization for C C CCC.
  2. Find a tangent vector to C C CCC at the point ( 4 , 2 ) ( 4 , 2 ) (4,2)(4,2)(4,2).

2.2 Cylindrical and spherical coordinates

There are several ways to specify the location of a point in 3 3 ^(3){ }^{3}3. The most common is to give the lengths of the projections onto the x , y x , y x-,y-x-, y-x,y and z z zzz-axes. These are, of course, the x x xxx-, y y yyy - and z z zzz-coordinates. We often call the ( x , y , z ) ( x , y , z ) (x,y,z)(x, y, z)(x,y,z) coordinate system
Fig. 2.1. The derivative of the parameterization φ ( t ) = ( t , t 2 ) φ ( t ) = t , t 2 varphi(t)=(t,t^(2))\varphi(t)=\left(t, t^{2}\right)φ(t)=(t,t2) is the vector 1 , 2 t 1 , 2 t (:1,2t:)\langle 1,2 t\rangle1,2t. When t = 1 t = 1 t=1t=1t=1 this is the vector 1 , 2 1 , 2 (:1,2:)\langle 1,2\rangle1,2, which we picture based at the point φ ( 1 ) = ( 1 , 1 ) φ ( 1 ) = ( 1 , 1 ) varphi(1)=(1,1)\varphi(1)=(1,1)φ(1)=(1,1).
Cartesian coordinates (after the mathematician René Descartes), or rectangular coordinates.
A second method of describing the location of a point is to use polar coordinates ( r , θ ) ( r , θ ) (r,theta)(r, \theta)(r,θ) to describe the projection onto the x y x y xyx yxy-plane, and the quantity z z zzz to describe the height off of the x y x y xyx yxy-plane (see
Figure 2.2). It follows that the relationships between r , θ , x r , θ , x r,theta,xr, \theta, xr,θ,x and y y yyy are the same as for polar coordinates:
x = r cos θ | r = x 2 + y 2 y = r sin θ | θ = tan 1 ( y x ) . . x = r cos θ r = x 2 + y 2 y = r sin θ θ = tan 1 y x . . [x=r cos theta|^(r)=sqrt(x^(2)+y^(2))],[y=r sin theta]|_(theta=tan^(-1)((y)/(x)).).\left.\begin{aligned} & x=\left.r \cos \theta\right|^{r}=\sqrt{x^{2}+y^{2}} \\ & y=r \sin \theta \end{aligned}\right|_{\theta=\tan ^{-1}\left(\frac{y}{x}\right) .} .x=rcosθ|r=x2+y2y=rsinθ|θ=tan1(yx)..
The ( r , θ , z r , θ , z r,theta,zr, \theta, zr,θ,z ) coordinates are called cylindrical co rdinates.
The third most common coordinate system is called spherical coordinates. In this system, one specifies the distance ρ ρ rho\rhoρ from the origin, the same angle θ θ theta\thetaθ from cylindrical coordinates and the angle φ φ varphi\varphiφ that a ray to the origin makes with the z z zzz-axis (see Figure 2.3). A little basic trigonometry yields the relationships:
\left.\begin{aligned} & x=\rho \sin \phi \cos \theta \\ & y=\rho \sin \phi \sin \theta \\ & z=\rho \cos \phi \end{aligned} \right\rvert\, \begin{aligned} & \rho=\sqrt{x^{2}+y^{2}+z^{2}} \\ & \phi=\tan ^{-1}\left(\frac{y}{x}\right) \\ & \left.\hline \frac{\sqrt{x^{2}+y^{2}}}{z}\right) . \end{aligned}Misplaced \hline
2.7. Find all of the relationships between the quantities r , θ r , θ r,thetar, \thetar,θ and z z zzz from cylindrical coordinates and the quantities ρ , θ ρ , θ rho,theta\rho, \thetaρ,θ and φ φ varphi\varphiφ from spherical coordinates.
Fig. 2.2. Cylindrical coordinates.
Fig. 2.3. Spherical coordinates.
Each coordinate system is useful for describing different graphs, as can be seen in the following examples.
Example 5. A cylinder of radius one, centered on the z z zzz-axis, can be described by equations in each coordinate system as follows:
  • Rectangular: x 2 + y 2 = 1 x 2 + y 2 = 1 x^(2)+y^(2)=1x^{2}+y^{2}=1x2+y2=1
  • Cylindrical: r = 1 r = 1 r=1r=1r=1
  • Spherical: ρ sin φ = 1 ρ sin φ = 1 rho sin varphi=1\rho \sin \varphi=1ρsinφ=1.
Example 6. A sphere of radius one is described by the equations:
  • Rectangular: x 2 + y 2 + z 2 = 1 x 2 + y 2 + z 2 = 1 x^(2)+y^(2)+z^(2)=1x^{2}+y^{2}+z^{2}=1x2+y2+z2=1
  • Cylindrical: r 2 + z 2 = 1 r 2 + z 2 = 1 r^(2)+z^(2)=1r^{2}+z^{2}=1r2+z2=1
  • Spherical: ρ = 1 ρ = 1 rho=1\rho=1ρ=1.
    2.8. Sketch the shape described by the following equations:
1. θ = π 4 . 2. z = r 2 . 3. ρ = ϕ . 4. ρ = cos ϕ . 5. r = cos θ . 6. z = r 2 1 . 7. z = r 2 + 1 . 8. r = θ .  1.  θ = π 4 .  2.  z = r 2 .  3.  ρ = ϕ .  4.  ρ = cos ϕ .  5.  r = cos θ .  6.  z = r 2 1 .  7.  z = r 2 + 1 .  8.  r = θ . {:[" 1. "theta=(pi)/(4).],[" 2. "z=r^(2).],[" 3. "rho=phi.],[" 4. "rho=cos phi.],[" 5. "r=cos theta.],[" 6. "z=sqrt(r^(2)-1).],[" 7. "z=sqrt(r^(2)+1).],[" 8. "r=theta.]:}\begin{aligned} & \text { 1. } \theta=\frac{\pi}{4} . \\ & \text { 2. } z=r^{2} . \\ & \text { 3. } \rho=\phi . \\ & \text { 4. } \rho=\cos \phi . \\ & \text { 5. } r=\cos \theta . \\ & \text { 6. } z=\sqrt{r^{2}-1} . \\ & \text { 7. } z=\sqrt{r^{2}+1} . \\ & \text { 8. } r=\theta . \end{aligned} 1. θ=π4. 2. z=r2. 3. ρ=ϕ. 4. ρ=cosϕ. 5. r=cosθ. 6. z=r21. 7. z=r2+1. 8. r=θ.
2.9. Find rectangular, cylindrical and spherical equations that describe the following shapes:
  1. A right, circular cone centered on the z z zzz-axis, with vertex at the origin.
  2. The x z x z xzx zxz-plane.
  3. The x y x y xyx yxy-plane.
  4. A plane that is at an angle of π 4 π 4 (pi)/(4)\frac{\pi}{4}π4 with both the x x xxx - and y y yyy-axes.
  5. The surface found by revolving the graph of z = x 3 z = x 3 z=x^(3)z=x^{3}z=x3 (where x x xxx 0 0 >= 0\geq 00 ) around the z z zzz-axis.

2.3 Parameterized surfaces in 3

A parameterization for a surface S S SSS in 3 3 ^(3){ }^{3}3 is a (one-to-one, onto, differentiable) function from some subset of R 2 R 2 R^(2)R^{2}R2 into 3 3 ^(3){ }^{3}3 whose image is S S SSS.
Example 7. The function φ ( u , u ) = ( u , u , 1 u 2 u 2 ) φ ( u , u ) = u , u , 1 u 2 u 2 varphi(u,u)=(u,u,sqrt()1-u^(2)-u^(2))\varphi(u, u)=\left(u, u, \sqrt{ } 1-u^{2}-u^{2}\right)φ(u,u)=(u,u,1u2u2), where ( u , u ) ( u , u ) (u,u)(u, u)(u,u) lies inside a disk of radius one, is a parameterization for the top half of the unit sphere.
One of the best ways to parameterize a surface is to find an equation in some coordinate system which can be used to eliminate one unknown coordinate. Then translate back to rectangular coordinates.
Example 8. An equation for the top half of the phere in cylindrical coordinates is r 2 + z 2 = 1 r 2 + z 2 = 1 r^(2)+z^(2)=1r^{2}+z^{2}=1r2+z2=1. Solving for z z zzz then gives us z = 1 r 2 z = 1 r 2 z=sqrt()1-r^(2)z=\sqrt{ } 1-r^{2}z=1r2. Translating to rectangular coordinates we have:
x = r cos θ y = r sin θ z = 1 r 2 . x = r cos θ y = r sin θ z = 1 r 2 . {:[x=r cos theta],[y=r sin theta],[],[z=sqrt(1-r^(2)).]:}\begin{aligned} & x=r \cos \theta \\ & y=r \sin \theta \\ & \\ & z=\sqrt{1-r^{2}} . \end{aligned}x=rcosθy=rsinθz=1r2.
Hence, a parameterization is given by he function
ϕ ( r , θ ) = ( r cos θ , r sin θ , 1 r 2 ) ϕ ( r , θ ) = r cos θ , r sin θ , 1 r 2 phi(r,theta)=(r cos theta,r sin theta,sqrt(1-r^(2)))\phi(r, \theta)=\left(r \cos \theta, r \sin \theta, \sqrt{1-r^{2}}\right)ϕ(r,θ)=(rcosθ,rsinθ,1r2)
where 0 r 1 0 r 1 0 <= r <= 10 \leq r \leq 10r1 and 0 θ 2 π 0 θ 2 π 0 <= theta <= 2pi0 \leq \theta \leq 2 \pi0θ2π.
Example 9. The equation ρ = φ ρ = φ rho=varphi\rho=\varphiρ=φ describes some surface in spherical coordinates. Translating to rectangular coordinates then gives us:
x = ρ sin ρ cos θ y = ρ sin ρ sin θ z = ρ cos ρ . x = ρ sin ρ cos θ y = ρ sin ρ sin θ z = ρ cos ρ . {:[x=rho sin rho cos theta],[y=rho sin rho sin theta],[z=rho cos rho.]:}\begin{gathered} x=\rho \sin \rho \cos \theta \\ y=\rho \sin \rho \sin \theta \\ z=\rho \cos \rho . \end{gathered}x=ρsinρcosθy=ρsinρsinθz=ρcosρ.
Hence, a parameterization for this surface is given by
ϕ ( ρ , θ ) = ( ρ sin ρ cos θ , ρ sin ρ sin θ , ρ cos ρ ) . ϕ ( ρ , θ ) = ( ρ sin ρ cos θ , ρ sin ρ sin θ , ρ cos ρ ) . phi(rho,theta)=(rho sin rho cos theta,rho sin rho sin theta,rho cos rho).\phi(\rho, \theta)=(\rho \sin \rho \cos \theta, \rho \sin \rho \sin \theta, \rho \cos \rho) .ϕ(ρ,θ)=(ρsinρcosθ,ρsinρsinθ,ρcosρ).
2.10. Find parameterizations of the surfaces described by the equations in Problem 2.8.
2.11. Find a parameterization for the graph of an equation of the form z = f ( x , y ) z = f ( x , y ) z=f(x,y)z=f(x, y)z=f(x,y).
2.12. Use the rectangular, cylindrical and spherical equations found in Problem 2.9 to parameterize the surfaces described there.
2.13. Use spherical coordinates to find a parameterization for the portion of the sphere of radius two, centered at the origin, which lies below the graph of z = r z = r z=rz=rz=r and above the x y x y xyx yxy-plane.
2.14. Sketch the surfaces given by the following parameterizations:
  1. ψ ( θ , ϕ ) = ( ϕ sin ϕ cos θ , ϕ sin ϕ sin θ , ϕ cos ϕ ) , 0 ϕ π 2 , 0 θ 2 π ψ ( θ , ϕ ) = ( ϕ sin ϕ cos θ , ϕ sin ϕ sin θ , ϕ cos ϕ ) , 0 ϕ π 2 , 0 θ 2 π psi(theta,phi)=(phi sin phi cos theta,phi sin phi sin theta,phi cos phi),0 <= phi <= (pi)/(2),0 <= theta <= 2pi\psi(\theta, \phi)=(\phi \sin \phi \cos \theta, \phi \sin \phi \sin \theta, \phi \cos \phi), 0 \leq \phi \leq \frac{\pi}{2}, 0 \leq \theta \leq 2 \piψ(θ,ϕ)=(ϕsinϕcosθ,ϕsinϕsinθ,ϕcosϕ),0ϕπ2,0θ2π.
  2. ϕ ( r , θ ) = ( r cos θ , r sin θ , cos r ) , 0 r 2 π , 0 θ 2 π ϕ ( r , θ ) = ( r cos θ , r sin θ , cos r ) , 0 r 2 π , 0 θ 2 π phi(r,theta)=(r cos theta,r sin theta,cos r),0 <= r <= 2pi,0 <= theta <= 2pi\phi(r, \theta)=(r \cos \theta, r \sin \theta, \cos r), 0 \leq r \leq 2 \pi, 0 \leq \theta \leq 2 \piϕ(r,θ)=(rcosθ,rsinθ,cosr),0r2π,0θ2π.
Just as we could differentiate parameterizations of curves in 2, we can also differentiate parameterizations of surfaces in 3 3 ^(3){ }^{3}3. In general, such a parameterization for a surface S S SSS can be written as
φ ( u , U ) = ( f ( u , U ) , g ( u , U ) , h ( u , U ) ) . φ ( u , U ) = ( f ( u , U ) , g ( u , U ) , h ( u , U ) ) . varphi(u,U)=(f(u,U),g(u,U),h(u,U)).\varphi(u, U)=(f(u, U), g(u, U), h(u, U)) .φ(u,U)=(f(u,U),g(u,U),h(u,U)).
Thus there are two variables we can differentiate with respect to: u u uuu and U U UUU. Each of these gives a vector which is tangent to the parameterized surface:
ϕ u = f u , g u , h u ϕ v = f v , g v , h v . ϕ u = f u , g u , h u ϕ v = f v , g v , h v . {:[(del phi)/(del u)=(:(del f)/(del u),(del g)/(del u),(del h)/(del u):)],[(del phi)/(del v)=(:(del f)/(del v),(del g)/(del v),(del h)/(del v):).]:}\begin{aligned} & \frac{\partial \phi}{\partial u}=\left\langle\frac{\partial f}{\partial u}, \frac{\partial g}{\partial u}, \frac{\partial h}{\partial u}\right\rangle \\ & \frac{\partial \phi}{\partial v}=\left\langle\frac{\partial f}{\partial v}, \frac{\partial g}{\partial v}, \frac{\partial h}{\partial v}\right\rangle . \end{aligned}ϕu=fu,gu,huϕv=fv,gv,hv.
The vectors ϕ u ϕ u (del phi)/()^((del )/(u))\frac{\partial \phi}{}{ }^{\frac{\partial}{u}}ϕu and ϕ v ϕ v (del phi)/(del v)\frac{\partial \phi}{\partial v}ϕv determine a plane which is tangent to the surface S S SSS at the point φ ( u , U ) φ ( u , U ) varphi(u,U)\varphi(u, \mathrm{U})φ(u,U).
2.15. Suppose some surface is described by the parameterization
ϕ ( u , v ) = ( 2 u , 3 v , u 2 + v 2 ) . ϕ ( u , v ) = 2 u , 3 v , u 2 + v 2 . phi(u,v)=(2u,3v,u^(2)+v^(2)).\phi(u, v)=\left(2 u, 3 v, u^{2}+v^{2}\right) .ϕ(u,v)=(2u,3v,u2+v2).
Find two (non-parallel) vectors which are tangent to this surface at the point ( 4 , 3 , 5 ) ( 4 , 3 , 5 ) (4,3,5)(4,3,5)(4,3,5).

2.4 Parameterized curves in 3

We begin with an example which demonstrates a parameterization of a curve in 3 3 ^(3){ }^{3}3.
Example 10. The function φ ( t ) = ( cos t , sin t , t ) φ ( t ) = ( cos t , sin t , t ) varphi(t)=(cos t,sin t,t)\varphi(t)=(\cos t, \sin t, t)φ(t)=(cost,sint,t) parameterizes a curve that spirals upward around a cylinder of radius one.
2.16. Describe the difference between the curves with the following parameterizations:
  1. ( cos t 2 , sin t 2 , t 2 ) cos t 2 , sin t 2 , t 2 (cos t^(2),sin t^(2),t^(2))\left(\cos t^{2}, \sin t^{2}, t^{2}\right)(cost2,sint2,t2).
  2. ( cos t , sin t , t 2 ) cos t , sin t , t 2 (cos t,sin t,t^(2))\left(\cos t, \sin t, t^{2}\right)(cost,sint,t2).
  3. ( t cos t , t sin t , t ) ( t cos t , t sin t , t ) (t cos t,t sin t,t)(t \cos t, t \sin t, t)(tcost,tsint,t).
  4. ( cos 1 t , sin 1 t , t ) cos 1 t , sin 1 t , t (cos^((1)/(t)),sin^((1)/(t)),t)\left(\cos ^{\frac{1}{t}}, \sin ^{\frac{1}{t}}, t\right)(cos1t,sin1t,t).
    2.17. Describe the lines given by the following parameterizations:
  5. ( t , 0 , 0 ) ( t , 0 , 0 ) (t,0,0)(t, 0,0)(t,0,0).
  6. ( 0 , 0 , t ) ( 0 , 0 , t ) (0,0,t)(0,0, t)(0,0,t).
  7. ( 0 , t , t ) ( 0 , t , t ) (0,t,t)(0, t, t)(0,t,t).
  8. ( t , t , t ) ( t , t , t ) (t,t,t)(t, t, t)(t,t,t).
In the previous section, we found parameterizations of surfaces by finding an equation for the surface (in some coordinate system), solving for a variable and then translating to rectangular coordinates. To find a parameterization of a curve in R 3 R 3 R^(3)\mathbb{R}^{3}R3, an effective strategy is to find some way to "eliminate" two coordinates (in some system), and then translate into rectangular coordinates. By "eliminating" a
coordinate we mean either expressing it as some constant, or expressing it as a function of the third, unknown coordinate.
Example 11. We demonstrate two ways to parameterize one of the lines that is at the intersection of the cone z 2 = x 2 + y 2 z 2 = x 2 + y 2 z^(2)=x^(2)+y^(2)z^{2}=x^{2}+y^{2}z2=x2+y2 and the plane y = 2 x y = 2 x y=2xy=2 xy=2x. The coordinate y y yyy is already expressed as a function of x x xxx. To express z z zzz as a function of x x xxx, we substitute 2 x 2 x 2x2 x2x for y y yyy in the first equation. This gives us z 2 = x 2 + ( 2 x ) 2 = 5 x 2 z 2 = x 2 + ( 2 x ) 2 = 5 x 2 z^(2)=x^(2)+(2x)^(2)=5x^(2)z^{2}=x^{2}+(2 x)^{2}=5 x^{2}z2=x2+(2x)2=5x2, or z = 5 x z = 5 x z=5xz=5 xz=5x (the negative root would give us the other intersection line). Hence, we get the paramaterization
ϕ ( x ) = ( x , 2 x , 5 x ) . ϕ ( x ) = ( x , 2 x , 5 x ) . phi(x)=(x,2x,sqrt5x).\phi(x)=(x, 2 x, \sqrt{5} x) .ϕ(x)=(x,2x,5x).
Another way to describe this line is with spherical coordinates. Note that for every point on the line φ = π 4 φ = π 4 varphi=(pi)/(4)\varphi=\frac{\pi}{4}φ=π4 (from the first equation) and θ θ theta\thetaθ = tan 1 2 = tan 1 2 =tan^(-1)2=\tan ^{-1} 2=tan12 (because tan θ = y / x = 2 tan θ = y / x = 2 tan theta=y//x=2\tan \theta=\mathrm{y} / \mathrm{x}=2tanθ=y/x=2, from the second equation). Converting to rectangular coordinates then gives us
ϕ ( ρ ) = ( ρ sin π 4 cos ( tan 1 2 ) , ρ sin π 4 sin ( tan 1 2 ) , ρ cos π 4 ) ϕ ( ρ ) = ρ sin π 4 cos tan 1 2 , ρ sin π 4 sin tan 1 2 , ρ cos π 4 phi(rho)=(rho sin((pi)/(4))cos(tan^(-1)2),rho sin((pi)/(4))sin(tan^(-1)2),rho cos((pi)/(4)))\phi(\rho)=\left(\rho \sin \frac{\pi}{4} \cos \left(\tan ^{-1} 2\right), \rho \sin \frac{\pi}{4} \sin \left(\tan ^{-1} 2\right), \rho \cos \frac{\pi}{4}\right)ϕ(ρ)=(ρsinπ4cos(tan12),ρsinπ4sin(tan12),ρcosπ4)
which simplifies to
ψ ( ρ ) = ( 10 ρ 10 , 10 ρ 5 , 2 ρ 2 ) . ψ ( ρ ) = 10 ρ 10 , 10 ρ 5 , 2 ρ 2 . psi(rho)=((sqrt10rho)/(10),(sqrt10rho)/(5),(sqrt2rho)/(2)).\psi(\rho)=\left(\frac{\sqrt{10} \rho}{10}, \frac{\sqrt{10} \rho}{5}, \frac{\sqrt{2} \rho}{2}\right) .ψ(ρ)=(10ρ10,10ρ5,2ρ2).
Note that dividing the first parameterization by 10 10 sqrt()10\sqrt{ } 1010 and simplifying yields the second parameterization.
2.18. Find a parameterization for the curve that is at the intersection of the plane x + y = 1 x + y = 1 x+y=1x+y=1x+y=1 and the cone z 2 = x 2 + y 2 z 2 = x 2 + y 2 z^(2)=x^(2)+y^(2)z^{2}=x^{2}+y^{2}z2=x2+y2.
2.19. Find two parameterizations for the circle that is at the intersection of the cylinder x 2 + y 2 = 4 x 2 + y 2 = 4 x^(2)+y^(2)=4x^{2}+y^{2}=4x2+y2=4 and the paraboloid z = x 2 + z = x 2 + z=x^(2)+z=x^{2}+z=x2+ y 2 y 2 y^(2)y^{2}y2.

2.5 Parameterized regions in 2 and

In Section 1.3, we learned how to integrate functions of multiple variables over rectangular regions. Eventually we will learn how to integrate such functions over regions of any shape. The trick will be to parameterize such regions by functions whose domain is a rectangle. Some cases of this are already familiar.
Example 12. A parameterization for the disk of radius one (that is, the set of points in 2 2 ^(2){ }^{2}2 which are at a distance of at most one from the origin) is given using polar coordinates:
ϕ ( r , θ ) = ( r cos θ , r sin θ ) , 0 r 1 , 0 θ 2 π . ϕ ( r , θ ) = ( r cos θ , r sin θ ) , 0 r 1 , 0 θ 2 π . phi(r,theta)=(r cos theta,r sin theta),0 <= r <= 1,0 <= theta <= 2pi.\phi(r, \theta)=(r \cos \theta, r \sin \theta), 0 \leq r \leq 1,0 \leq \theta \leq 2 \pi .ϕ(r,θ)=(rcosθ,rsinθ),0r1,0θ2π.
2.20. Let B B BBB be the ball of radius one in (i.e., the set of points satisfying x 2 + y 2 + z 2 1 x 2 + y 2 + z 2 1 x^(2)+y^(2)+z^(2) <= 1x^{2}+y^{2}+z^{2} \leq 1x2+y2+z21 ).
  1. Use spherical coordinates to find a parameterization for B B BBB.
  2. Find a parameterization for the intersection of B B BBB with the first octant.
    2.21. The "solid cylinder" of height one and radius r r rrr in 3 3 ^(3){ }^{3}3 is the set of points inside the cylinder x 2 + y 2 = r 2 x 2 + y 2 = r 2 x^(2)+y^(2)=r^(2)x^{2}+y^{2}=r^{2}x2+y2=r2, and between the planes z z zzz = 0 = 0 =0=0=0 and z = 1 z = 1 z=1z=1z=1.
  3. Use cylindrical coordinates to find a parameterization for the solid cylinder of height one and radius one.
  4. Find a parameterization for the region that is inside the solid cylinder of height one and radius two and outside the cylinder of radius one.
Example 13. A common type of region to integrate over is one that is bounded by the graphs of two functions. Suppose R R RRR is the region in 2 2 ^(2){ }^{2}2 above the graph of y = g 1 ( x ) y = g 1 ( x ) y=g_(1)(x)y=g_{1}(x)y=g1(x), below the graph of y = g 2 ( x ) y = g 2 ( x ) y=g_(2)(x)y=g_{2}(x)y=g2(x) and between the lines x = a x = a x=ax=ax=a and x = b x = b x=bx=bx=b. A parameterization for R R RRR (check this!) is given by
ϕ ( x , t ) = ( x , tg 2 ( x ) + ( 1 t ) g 1 ( x ) ) , a x b , 0 t 1 . ϕ ( x , t ) = x , tg 2 ( x ) + ( 1 t ) g 1 ( x ) , a x b , 0 t 1 . phi(x,t)=(x,tg_(2)(x)+(1-t)g_(1)(x)),a <= x <= b,0 <= t <= 1.\phi(x, t)=\left(x, \operatorname{tg}_{2}(x)+(1-t) g_{1}(x)\right), a \leq x \leq b, 0 \leq t \leq 1 .ϕ(x,t)=(x,tg2(x)+(1t)g1(x)),axb,0t1.
2.22. Let R R RRR be the region between the (polar) graphs of r = f 1 ( θ ) r = f 1 ( θ ) r=f_(1)(theta)r=f_{1}(\theta)r=f1(θ) and r = f 2 ( θ ) r = f 2 ( θ ) r=f_(2)(theta)r=f_{2}(\theta)r=f2(θ), where a θ b a θ b a <= theta <= ba \leq \theta \leq baθb. Find a parameterization for R R RRR.
2.23. Find a parameterization for the region in 2 2 ^(2){ }^{2}2 bounded by the ellipse whose x x xxx-intercepts are 3 and -3 and y y yyy-intercepts are 2 and -2. (Hint. Start with the parameterization given in Example 12.)
2.24. Sketch the region in 2 2 ^(2){ }^{2}2 parameterized by the following:
ϕ ( r , θ ) = ( 2 r cos θ , r sin θ ) ϕ ( r , θ ) = ( 2 r cos θ , r sin θ ) phi(r,theta)=(2r cos theta,r sin theta)\phi(r, \theta)=(2 r \cos \theta, r \sin \theta)ϕ(r,θ)=(2rcosθ,rsinθ)
where 1 r 2 1 r 2 1 <= r <= 21 \leq r \leq 21r2 and 0 θ π / 2 0 θ π / 2 0 <= theta <= pi//20 \leq \theta \leq \pi / 20θπ/2.

3

Introduction to Forms

3.1 So what is a differential form?

A differential form is simply this: an integrand. In other words, it is a thing which can be integrated over some (often complicated) domain. For example, consider the following integral:0 x 2 d x x 2 d x x^(2)dxx^{2} d xx2dx. This notation indicates that we are integrating x 2 x 2 x^(2)x^{2}x2 over the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1]. In this case, x 2 d x x 2 d x x^(2)dxx^{2} d xx2dx is a differential form. If you have had no exposure to this subject this may make you a little uncomfortable. After all, in calculus we are taught that x 2 x 2 x^(2)x^{2}x2 is the integrand. The symbol " d x d x dx^(')d x^{\prime}dx is only there to delineate when the integrand has ended and what variable we are integrating with respect to. However, as an object in itself, we are not taught any meaning for " d x d x dxd xdx." Is it a function? Is it an operator on functions? Some professors call it an "infinitesimal" quantity. This is very tempting. After all, 0 x 2 d x 0 x 2 d x 0x^(2)dx0 x^{2} d x0x2dx is defined to be the limit, as n n n rarr oon \rightarrow \inftyn, off i = 1 x 2 n Δ x i = 1 x 2 n Δ x sum_(i=1x^(2))^(n)Delta x\sum_{i=1 x^{2}}^{n} \Delta xi=1x2nΔx, where { x i } x i {x_(i)}\left\{x_{i}\right\}{xi} are n n nnn evenly spaced points in the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1], and Δ x = i n Δ x = i n Delta x=in\Delta x=i nΔx=in. When we take the limit, the symbol " Σ Σ Sigma\SigmaΣ " becomes " int\int," and the symbol " Δ x Δ x Delta x\Delta xΔx " becomes " d x d x dxd xdx." This implies that d x = lim Δ x 0 Δ x d x = lim Δ x 0 Δ x dx=lim_(Delta_(x rarr0))Delta xd x=\lim _{\Delta_{x \rightarrow 0}} \Delta xdx=limΔx0Δx, which is absurd. lim Δ x 0 Δ x = 0 ! ! lim Δ x 0 Δ x = 0 ! ! lim_(Delta_(x rarr0))Delta x=0!!\lim _{\Delta_{x \rightarrow 0}} \Delta x=0!!limΔx0Δx=0!! We are not trying to make the argument that the symbol " d x d x dxd xdx " should be eliminated. It does have meaning. This is one of the many mysteries that this book will reveal.
One word of caution here: not all integrands are differential forms. In fact, in the appendix we will see how to calculate arc length and surface area. These calculations involve integrands which are not differential forms. Differential forms are simply natural objects to
integrate, and also the first that one should study. As we shall see, this is much like beginning the study of all functions by understanding linear functions. The naive student may at first object to this, since linear functions are a very restrictive class. On the other hand, eventually we learn that any differentiable function (a much more general class) can be locally approximated by a linear function. Hence, in some sense, the linear functions are the most important ones. In the same way, one can make the argument that differential forms are the most important integrands.

3.2 Generalizing the integral

Let's begin by studying a simple example, and trying to figure out how and what to integrate. The function f ( x , y ) = y 2 f ( x , y ) = y 2 f(x,y)=y^(2)f(x, y)=y^{2}f(x,y)=y2 maps to Let M M MMM denote the top half of the circle of radius one, centered at the origin. Let's restrict the function f f fff to the domain, M M MMM, and try to integrate it. Here we encounter our first problem: The given description of M M MMM is not particularly useful. If M M MMM were something more complicated, it would have been much harder to describe it in words as we have just done. A parameterization is far easier to communicate, and far easier to use to determine which points of 2 2 ^(2){ }^{2}2 are elements of M M MMM, and which are not. But there are lots of parameterizations of M M MMM. Here are two which we shall use:
ϕ 1 ( a ) = ( a , 1 a 2 ) , where 1 a 1 , ϕ 1 ( a ) = a , 1 a 2 , where  1 a 1 phi_(1)(a)=(a,sqrt(1-a^(2)))", where "-1 <= a <= 1", "\phi_{1}(a)=\left(a, \sqrt{1-a^{2}}\right) \text {, where }-1 \leq a \leq 1 \text {, }ϕ1(a)=(a,1a2), where 1a1
and
ϕ 2 ( t ) = ( cos ( t ) , sin ( t ) ) , where 0 t π . ϕ 2 ( t ) = ( cos ( t ) , sin ( t ) ) , where  0 t π phi_(2)(t)=(cos(t),sin(t))", where "0 <= t <= pi". "\phi_{2}(t)=(\cos (t), \sin (t)) \text {, where } 0 \leq t \leq \pi \text {. }ϕ2(t)=(cos(t),sin(t)), where 0tπ
Here is the trick: integrating f f fff over M M MMM is difficult. It may not even be clear as to what this means. But perhaps we can use Ψ 1 Ψ 1 Psi_(1)\Psi_{1}Ψ1 to translate this problem into an integral over the interval [-1, 1]. After all, an integral is a big sum. If we add up all the numbers f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) for all the points, ( x , y ) ( x , y ) (x,y)(x, y)(x,y), of M M MMM, shouldn't we get the same thing as if we
added up all the numbers f ( φ 1 ( a ) ) f φ 1 ( a ) f(varphi_(1)(a))f\left(\varphi_{1}(a)\right)f(φ1(a)), for all the points, a a aaa, of [ 1 , 1 ] [ 1 , 1 ] [-1,1][-1,1][1,1] (see Fig. 3.1)?
Fig. 3.1. Shouldn't the integral of f f fff over M M MMM be the same as the integral of f v φ f v φ f_(v)varphif_{\mathrm{v}} \varphifvφ over [-1, 1]?
Let's try it. Φ 1 ( a ) = ( a , 1 a 2 ) Φ 1 ( a ) = a , 1 a 2 Phi_(1)(a)=(a,sqrt(1-a^(2)))\Phi_{1}(a)=\left(a, \sqrt{1-a^{2}}\right)Φ1(a)=(a,1a2), so f ( φ 1 ( a ) ) = 1 a 2 f φ 1 ( a ) = 1 a 2 f(varphi_(1)(a))=1-a^(2)f\left(\varphi_{1}(a)\right)=1-a^{2}f(φ1(a))=1a2. Hence, we are saying that the integral of f f fff over M M MMM should be the same as-1(1 a 2 ) d a a 2 d a {:a^(2))da\left.a^{2}\right) d aa2)da. Using -1 a little calculus, we can determine that this evaluates to 4 / 3 4 / 3 4//34 / 34/3
Let's try this again, this time using 2 2 _(2)~_{2} 2. Using the same argument, the integral of f f fff over M M MMM should be the same aso f ( φ 2 ( t ) ) d t = 0 0 f φ 2 ( t ) d t = 0 0 f(varphi_(2)(t))dt=0int_(0)f\left(\varphi_{2}(t)\right) d t=0 \int_{0}f(φ2(t))dt=00 sin 2 ( t ) d t = π / 2 sin 2 ( t ) d t = π / 2 sin^(2)(t)dt=pi//2\sin ^{2}(t) d t=\pi / 2sin2(t)dt=π/2.
But hold on! The problem was stated before any parameterizations were chosen. Shouldn't the answer be independent of which one was picked? It would not be a very meaningful problem if two people could get different correct answers, depending on how they went about solving it. Something strange is going on!

3.3 Interlude: a review of single variable integration

In order to understand what happened, we must first review the definition of the Riemann integral. In the usual definition of the integral the first step is to divide b b bbb the interval up into n n nnn evenly spaced subintervals. Thus, a f ( x ) d x a f ( x ) d x int_(a)f(x)dx\int_{\mathrm{a}} f(x) d xaf(x)dx is defined to be the limit, as n n nnn rarr oo\rightarrow \infty, of i = 1 n f ( x i ) Δ x i = 1 n f x i Δ x sum_(i=1)^(n)f(x_(i))Delta x\sum_{i=1}^{n} f\left(x_{i}\right) \Delta xi=1nf(xi)Δx, where { x i } x i {x_(i)}\left\{x_{i}\right\}{xi} are n n nnn evenly spaced points in the interval [ a , b ] [ a , b ] [a,b][a, b][a,b], and Δ i = ( b a ) / n Δ i = ( b a ) / n Delta i=(b-a)//n\Delta i=(b-a) / nΔi=(ba)/n. But what if the points { x i } x i {x_(i)}\left\{x_{i}\right\}{xi} are
not n n nnn evenly spaced? We can still write down a reasonable sum: i = 1 i = 1 sum_(i=1)\sum_{i=1}i=1 f ( x i ) Δ x i j f x i Δ x i j f(x_(i))Deltax_(ij)f\left(x_{i}\right) \Delta x_{i j}f(xi)Δxij where now Δ x i = x i + 1 x i Δ x i = x i + 1 x i Deltax_(i)=x_(i+1)-x_(i)\Delta x_{i}=x_{i+1}-x_{i}Δxi=xi+1xi. In order to make the integral well-defined, we can no longer take the limit as n n n rarr oon \rightarrow \inftyn. Instead, we must let max { Δ x i } 0 max Δ x i 0 max{Deltax_(i)}rarr0\max \left\{\Delta x_{i}\right\} \rightarrow 0max{Δxi}0. It is a basic result of analysis that if this limit converges, then it does not matter how we picked the points { x i x i {x_(i):}\left\{x_{i}\right.{xi }; the limit will converge to the same number. It is this b b bbb number that we define to be the value of a f ( x ) d x a f ( x ) d x af(x)dxa f(x) d xaf(x)dx.

3.4 What went wrong?

We are now ready to figure out what happened in Section 3.2. Obviously, 1 f ( φ 1 ( a ) ) d a 1 f φ 1 ( a ) d a -1f(varphi_(1)(a))da-1 f\left(\varphi_{1}(a)\right) d a1f(φ1(a))da was not what we wanted. But let's not give up on our general approach just yet; it would still be great if we could use Φ 1 Φ 1 Phi_(1)\Phi_{1}Φ1 to find some function that we can integrate on [ 1 , 1 ] [ 1 , 1 ] [-1,1][-1,1][1,1] that will give us the same answer as the integral of f f fff over M M MMM. For now, let's call this mystery function " F ( a ) F ( a ) F(a)F(a)F(a)."
Let's look at the Riemann sum that we get for-1F(a)da, when we divide the interval up into n n nnn pieces, each of width Δ a : i = 1 F ( a j ) Δ a Δ a : i = 1 F a j Δ a Delta a:^(i=1)F(a_(j))Delta a\Delta a:{ }^{i=1} F\left(a_{j}\right) \Delta aΔa:i=1F(aj)Δa. Examine Figure 3.2 to see what happens to the points, a i a i a_(i)a_{i}ai, under th unction, Φ 1 Φ 1 Phi_(1)\Phi_{1}Φ1. Notice that the points { φ 1 ( a i ) } φ 1 a i {varphi_(1)(a_(i))}\left\{\varphi_{1}\left(a_{i}\right)\right\}{φ1(ai)} are not evenly spaced along M M MMM. To use these points to estimate the integral of f f fff over M M MMM, we would have to use the approach from the previous n n nnn section. A Riemann sum for f f fff over M M MMM would be f ( φ 1 ( a i ) ) l i f φ 1 a i l i sum f(varphi_(1)(a_(i)))l_(i)\sum f\left(\varphi_{1}\left(a_{i}\right)\right) l_{i}f(φ1(ai))li, where the l i i l i i l_(i)il_{i} ilii represent the arc length, along M M MMM, between φ 1 ( a i ) φ 1 a i varphi_(1)(a_(i))\varphi_{1}\left(a_{i}\right)φ1(ai) a φ 1 ( a i + 1 ) φ 1 a i + 1 varphi_(1)(a_(i+1))\varphi_{1}\left(a_{i+1}\right)φ1(ai+1).
Fig. 3.2. We want i = 1 n F ( a i ) Δ a = i = 1 n f ( φ 1 ( a i ) ) L i i = 1 n F a i Δ a = i = 1 n f φ 1 a i L i sum_(i=1)^(n)F(a_(i))Delta a=sum_(i=1)^(n)f(varphi_(1)(a_(i)))L_(i)\sum_{i=1}^{n} F\left(a_{i}\right) \Delta a=\sum_{i=1}^{n} f\left(\varphi_{1}\left(a_{i}\right)\right) L_{i}i=1nF(ai)Δa=i=1nf(φ1(ai))Li.
This is a bit problematic, however, since arc-length is generally hard to calculate. Instead, we can approximate l i l i l_(i)l_{i}li by substituting in the length of the line segment which connects φ 1 ( a i ) φ 1 a i varphi_(1)(a_(i))\varphi_{1}\left(a_{i}\right)φ1(ai) to φ 1 ( a i + 1 ) φ 1 a i + 1 varphi_(1)(a_(i+1))\varphi_{1}\left(a_{i+1}\right)φ1(ai+1), which we shall denote as L i L i L_(i)L_{i}Li. Note that this approximation gets better and better as we let n n n rarr oon \rightarrow \inftyn. Hence, when we take the limit, it does not matter if we use l i l i l_(i)l_{i}li or L i L i L_(i)L_{i}Li.
So our goal is to find a function, F ( a ) F ( a ) F(a)F(a)F(a), on the interval [ 1 , 1 ] [ 1 , 1 ] [-1,1][-1,1][1,1], so that
i = 1 n F ( a i ) Δ a = i = 1 n f ( ϕ 1 ( a i ) ) L i . i = 1 n F a i Δ a = i = 1 n f ϕ 1 a i L i . sum_(i=1)^(n)F(a_(i))Delta a=sum_(i=1)^(n)f(phi_(1)(a_(i)))L_(i).\sum_{i=1}^{n} F\left(a_{i}\right) \Delta a=\sum_{i=1}^{n} f\left(\phi_{1}\left(a_{i}\right)\right) L_{i} .i=1nF(ai)Δa=i=1nf(ϕ1(ai))Li.
Of course this equality will hold if F ( a i ) Δ a = f ( φ 1 ( a i ) ) L i F a i Δ a = f φ 1 a i L i F(a_(i))Delta a=f(varphi_(1)(a_(i)))L_(i)F\left(a_{i}\right) \Delta a=f\left(\varphi_{1}\left(a_{i}\right)\right) L_{i}F(ai)Δa=f(φ1(ai))Li. Solving, we get F ( a i ) = f ( ϕ 1 ( a i ) ) L i Δ a F a i = f ϕ 1 a i L i Δ a ^(F)(a_(i))=(f(phi_(1)(a_(i)))L_(i))/(Delta a)^{F}\left(a_{i}\right)=\frac{f\left(\phi_{1}\left(a_{i}\right)\right) L_{i}}{\Delta a}F(ai)=f(ϕ1(ai))LiΔa What happens to this function as Δ a 0 Δ a 0 Delta a rarr0\Delta a \rightarrow 0Δa0 ? First, note that L i = | φ 1 ( a i + 1 ) φ 1 ( a i ) | L i = φ 1 a i + 1 φ 1 a i L_(i)=|varphi_(1)(a_(i+1))-varphi_(1)(a_(i))|L_{i}=\left|\varphi_{1}\left(a_{i+1}\right)-\varphi_{1}\left(a_{i}\right)\right|Li=|φ1(ai+1)φ1(ai)|. Hence,
lim Δ a 0 F ( a i ) = lim Δ a 0 f ( ϕ 1 ( a i ) ) L i Δ a = lim Δ a 0 f ( ϕ 1 ( a i ) ) | ϕ 1 ( a i + 1 ) ϕ 1 ( a i ) | Δ a = f ( ϕ 1 ( a i ) ) lim Δ a 0 | ϕ 1 ( a i + 1 ) ϕ 1 ( a i ) | Δ a = f ( ϕ 1 ( a i ) ) | lim Δ a 0 ϕ 1 ( a i + 1 ) ϕ 1 ( a i ) Δ a | lim Δ a 0 F a i = lim Δ a 0 f ϕ 1 a i L i Δ a = lim Δ a 0 f ϕ 1 a i ϕ 1 a i + 1 ϕ 1 a i Δ a = f ϕ 1 a i lim Δ a 0 ϕ 1 a i + 1 ϕ 1 a i Δ a = f ϕ 1 a i lim Δ a 0 ϕ 1 a i + 1 ϕ 1 a i Δ a {:[lim_(Delta a rarr0)F(a_(i))=lim_(Delta a rarr0)(f(phi_(1)(a_(i)))L_(i))/(Delta a)],[=lim_(Delta a rarr0)(f(phi_(1)(a_(i)))|phi_(1)(a_(i+1))-phi_(1)(a_(i))|)/(Delta a)],[=f(phi_(1)(a_(i)))lim_(Delta a rarr0)(|phi_(1)(a_(i+1))-phi_(1)(a_(i))|)/(Delta a)],[=f(phi_(1)(a_(i)))|lim_(Delta a rarr0)(phi_(1)(a_(i+1))-phi_(1)(a_(i)))/(Delta a)|]:}\begin{aligned} \lim _{\Delta a \rightarrow 0} F\left(a_{i}\right) & =\lim _{\Delta a \rightarrow 0} \frac{f\left(\phi_{1}\left(a_{i}\right)\right) L_{i}}{\Delta a} \\ & =\lim _{\Delta a \rightarrow 0} \frac{f\left(\phi_{1}\left(a_{i}\right)\right)\left|\phi_{1}\left(a_{i+1}\right)-\phi_{1}\left(a_{i}\right)\right|}{\Delta a} \\ & =f\left(\phi_{1}\left(a_{i}\right)\right) \lim _{\Delta a \rightarrow 0} \frac{\left|\phi_{1}\left(a_{i+1}\right)-\phi_{1}\left(a_{i}\right)\right|}{\Delta a} \\ & =f\left(\phi_{1}\left(a_{i}\right)\right)\left|\lim _{\Delta a \rightarrow 0} \frac{\phi_{1}\left(a_{i+1}\right)-\phi_{1}\left(a_{i}\right)}{\Delta a}\right| \end{aligned}limΔa0F(ai)=limΔa0f(ϕ1(ai))LiΔa=limΔa0f(ϕ1(ai))|ϕ1(ai+1)ϕ1(ai)|Δa=f(ϕ1(ai))limΔa0|ϕ1(ai+1)ϕ1(ai)|Δa=f(ϕ1(ai))|limΔa0ϕ1(ai+1)ϕ1(ai)Δa|
But lim Δ a 0 ϕ 1 ( a i + 1 ) ϕ 1 ( a i ) Δ a lim Δ a 0 ϕ 1 a i + 1 ϕ 1 a i Δ a lim Delta a rarr0(phi_(1)(a_(i+1))-phi_(1)(a_(i)))/(Delta a)\lim \Delta a \rightarrow 0 \frac{\phi_{1}\left(a_{i+1}\right)-\phi_{1}\left(a_{i}\right)}{\Delta a}limΔa0ϕ1(ai+1)ϕ1(ai)Δa is precisely the definition of the derivative of φ 1 φ 1 varphi_(1)\varphi_{1}φ1 at a i j d ϕ 1 d a ( a i ) a i j d ϕ 1 d a a i a_(ij)(dphi_(1))/(da)(a_(i))a_{i j} \frac{d \phi_{1}}{d a}\left(a_{i}\right)aijdϕ1da(ai). Hence, we have lim a 0 F ( a i ) = f ( ϕ 1 ( a i ) ) | d ϕ 1 d a ( a i ) | a 0 F a i = f ϕ 1 a i d ϕ 1 d a a i /_\a rarr0F(a_(i))=f(phi_(1)(a_(i)))|(dphi_(1))/(da)(a_(i))|\triangle a \rightarrow 0 F\left(a_{i}\right)=f\left(\phi_{1}\left(a_{i}\right)\right)\left|\frac{d \phi_{1}}{d a}\left(a_{i}\right)\right|a0F(ai)=f(ϕ1(ai))|dϕ1da(ai)|. Finally, this means that the is 1 f ( ϕ 1 ( a ) ) | d ϕ 1 d a | d a is  1 f ϕ 1 ( a ) d ϕ 1 d a d a int_("is ")^(1)f(phi_(1)(a))|(dphi_(1))/(da)|da\int_{\text {is }}^{1} f\left(\phi_{1}(a)\right)\left|\frac{d \phi_{1}}{d a}\right| d ais 1f(ϕ1(a))|dϕ1da|da integral we want to compute is- 1
1 1 1 1 f ( ϕ 1 ( a ) ) | d ϕ 1 d a | d a = 0 π f ( ϕ 2 ( t ) ) | d ϕ 2 d t | d t 1 1 1 1 f ϕ 1 ( a ) d ϕ 1 d a d a = 0 π f ϕ 2 ( t ) d ϕ 2 d t d t int_(-1)^(1)int_(-1)^(1)f(phi_(1)(a))|(dphi_(1))/(da)|da=int_(0)^(pi)f(phi_(2)(t))|(dphi_(2))/(dt)|dt\int_{-1}^{1} \int_{-1}^{1} f\left(\phi_{1}(a)\right)\left|\frac{d \phi_{1}}{d a}\right| d a=\int_{0}^{\pi} f\left(\phi_{2}(t)\right)\left|\frac{d \phi_{2}}{d t}\right| d t1111f(ϕ1(a))|dϕ1da|da=0πf(ϕ2(t))|dϕ2dt|dt , using the function, f f fff, defined in Section 3.2.
Recall that d ϕ 1 d a d ϕ 1 d a (dphi_(1))/(da)\frac{d \phi_{1}}{d a}dϕ1da is a vector, based at the point φ φ varphi\varphiφ (a), tangent to M M MMM. If we think of a a aaa as a time parameter, then the length of d ϕ 1 d a d ϕ 1 d a (dphi_(1))/(da)\frac{d \phi_{1}}{d a}dϕ1da tells us how fast φ 1 ( a ) φ 1 ( a ) varphi_(1)(a)\varphi_{1}(a)φ1(a) is moving along M M MMM. How can we generalize the integral, 1 1 f ( ϕ 1 ( a ) ) | d ϕ 1 d a | 1 1 f ϕ 1 ( a ) d ϕ 1 d a int_(-1)^(1)f(phi_(1)(a))|(dphi_(1))/(da)|\int_{-1}^{1} f\left(\phi_{1}(a)\right)\left|\frac{d \phi_{1}}{d a}\right|11f(ϕ1(a))|dϕ1da|
1 d a 1 d a -1quad da-1 \quad d a1da ? Note that the bars | | | | |*||\cdot||| denote a function that "eats" vectors and "spits out" real numbers. So we can generalize the integral by looking at other such functions. In other words, a
1 1 f ( ϕ 1 ( a ) ) ω ( d ϕ 1 d a ) d a 1 1 f ϕ 1 ( a ) ω d ϕ 1 d a d a int_(-1)^(1)f(phi_(1)(a))omega((dphi_(1))/(da))da\int_{-1}^{1} f\left(\phi_{1}(a)\right) \omega\left(\frac{d \phi_{1}}{d a}\right) d a11f(ϕ1(a))ω(dϕ1da)da
more general integral would be-1 function of points and ω ω omega\omegaω is a function of vec ors.
It is not the purpose of the present work to undertake a study of integrating with respect to all possible functions, ω ω omega\omegaω. However, as with the study of functions of real variables, a natural place to start is with linear functions. This is the study of differential forms. A differential form is precisely a linear function which eats vectors, spits out numbers and is used in integration. The strength of differential forms lies in the fact that their integrals do not depend on a choice of parameterization.

3.5 What about surfaces?

Let's repeat the previous discussion (faster this time), bumping everything up a dimension. Let f : 3 f : 3 f:^(3)rarrf:{ }^{3} \rightarrowf:3 be given by f ( x , y , z ) = f ( x , y , z ) = f(x,y,z)=f(x, y, z)=f(x,y,z)= z 2 z 2 z^(2)z^{2}z2. Let M M MMM be the top half of the sphere of radius one, centered at the origin. We can parameterize M M MMM by the function, φ φ varphi\varphiφ, where φ ( r , θ ) = ( r φ ( r , θ ) = ( r varphi(r,theta)=(r\varphi(r, \theta)=(rφ(r,θ)=(r cos ( θ ) , r sin ( θ ) , 1 r 2 ) , 0 r 1 cos ( θ ) , r sin ( θ ) , 1 r 2 , 0 r 1 {: cos(theta),r sin(theta),sqrt(1-r^(2))),0 <= r <= 1\left.\cos (\theta), r \sin (\theta), \sqrt{1-r^{2}}\right), 0 \leq r \leq 1cos(θ),rsin(θ),1r2),0r1, and 0 θ 2 π 0 θ 2 π 0 <= theta <= 2pi0 \leq \theta \leq 2 \pi0θ2π. Again, our goal is not to figure out how to actually integrate f f fff over M M MMM, but to use φ φ varphi\varphiφ to set up an equivalent integral over the rectangle, R = [ 0 , 1 ] R = [ 0 , 1 ] R=[0,1]R=[0,1]R=[0,1] × [ 0 , 2 π ] × [ 0 , 2 π ] xx[0,2pi]\times[0,2 \pi]×[0,2π].
Let { x i , j } x i , j {x_(i,j)}\left\{x_{i, j}\right\}{xi,j} be a lattice of evenly spaced points in R R RRR. Let Δ r = x i + 1 , j Δ r = x i + 1 , j Delta r=x_(i+1,j)^(-)\Delta r=x_{i+1, j}{ }^{-}Δr=xi+1,j x i , j x i , j x_(i,j)x_{i, j}xi,j, and Δ θ = x i , j + 1 x i , j Δ θ = x i , j + 1 x i , j Delta theta=x_(i,j+1)-x_(i,j)\Delta \theta=x_{i, j+1}-x_{i, j}Δθ=xi,j+1xi,j. By definition, the integral over R R RRR of a function, F ( x ) F ( x ) F(x)F(x)F(x), is equal to lim Δ r Δ θ 0 Σ F ( x i , j ) Δ r Δ θ lim Δ r Δ θ 0 Σ F x i , j Δ r Δ θ lim_(Delta r)Deltatheta_(rarr0)Sigma F(x_(i,j))Delta r Delta theta\lim _{\Delta r} \Delta \theta_{\rightarrow 0} \Sigma F\left(x_{i, j}\right) \Delta r \Delta \thetalimΔrΔθ0ΣF(xi,j)ΔrΔθ.
To use the mesh o points, φ ( x i , j ) φ x i , j varphi(x_(i,j))\varphi\left(x_{i, j}\right)φ(xi,j), in M M MMM to set up a Riemann sum, we write down the following sum: f ( φ ( x i , j ) ) f φ x i , j sum f(varphi(x_(i,j)))\sum f\left(\varphi\left(x_{i, j}\right)\right)f(φ(xi,j)) Area ( L i , j ) L i , j (L_(i,j))\left(L_{i, j}\right)(Li,j), where L i , j L i , j L_(i,j)L_{i, j}Li,j is the rectangle spanned by the vectors φ ( x i + 1 , j ) φ ( x i , j ) φ x i + 1 , j φ x i , j varphi(x_(i+1,j))-varphi(x_(i,j))\varphi\left(x_{i+1, j}\right)-\varphi\left(x_{i, j}\right)φ(xi+1,j)φ(xi,j) and φ ( x i , j + 1 ) φ x i , j + 1 varphi(x_(i,j+1))\varphi\left(x_{i, j+1}\right)φ(xi,j+1) φ ( x i , j ) φ x i , j varphi(x_(i,j))\varphi\left(x_{i, j}\right)φ(xi,j). If we want our Riemann sum over R R RRR to equal this sum, then we end up with F ( x i , j ) = f ( ϕ ( x i , j ) ) Area ( L i , j ) Δ r Δ θ F x i , j = f ϕ x i , j Area L i , j Δ r Δ θ F(x_(i,j))=(f(phi(x_(i,j)))Area(L_(i,j)))/(Delta r Delta theta)F\left(x_{i, j}\right)=\frac{f\left(\phi\left(x_{i, j}\right)\right) \operatorname{Area}\left(L_{i, j}\right)}{\Delta r \Delta \theta}F(xi,j)=f(ϕ(xi,j))Area(Li,j)ΔrΔθ.
Fig. 3.3. Setting up the Riemann sum for the integral of z 2 z 2 z^(2)z^{2}z2 over the top half of the sphere of radius one.
We now leave it as an exercise to show that as Δ r Δ r Delta r\Delta rΔr and Δ θ Δ θ Delta theta\Delta \thetaΔθ get small, Area ( L i , j ) Δ r Δ θ  Area  L i , j Δ r Δ θ (" Area "(L_(i,j)))/(Delta r Delta theta)\frac{\text { Area }\left(L_{i, j}\right)}{\Delta r \Delta \theta} Area (Li,j)ΔrΔθ converges to the area of the parallelogram spanned by the vectors ϕ r ( x i , j ) ϕ r x i , j (del phi)/(del r)(x_(i,j))\frac{\partial \phi}{\partial r}\left(x_{i, j}\right)ϕr(xi,j) and ϕ θ ( x i , j ) ϕ θ x i , j (del phi)/(del theta)(x_(i,j))\frac{\partial \phi}{\partial \theta}\left(x_{i, j}\right)ϕθ(xi,j). The upshot of all this is that the integral we want to evaluate is the following:
R f ( ϕ ( r , θ ) ) Area ( ϕ r , ϕ θ ) d r d θ R f ( ϕ ( r , θ ) ) Area ϕ r , ϕ θ d r d θ int_(R)f(phi(r,theta))Area((del phi)/(del r),(del phi)/(del theta))drd theta\int_{R} f(\phi(r, \theta)) \operatorname{Area}\left(\frac{\partial \phi}{\partial r}, \frac{\partial \phi}{\partial \theta}\right) d r d \thetaRf(ϕ(r,θ))Area(ϕr,ϕθ)drdθ
3.2. Compute the value of this integral for the function f ( x , y , z ) = f ( x , y , z ) = f(x,y,z)=f(x, y, z)=f(x,y,z)= z 2 z 2 z^(2)z^{2}z2.
The point of all this is not the specific integral that we have arrived at, but the form of the integral. We integrate f v φ f v φ f_(v)varphif_{\mathrm{v}} \varphifvφ (as in the previous section), times a function which takes two vectors and returns a real number. Once again, we can generalize this by using other such functions:
R f ( ϕ ( r , θ ) ) ω ( ϕ r , ϕ θ ) d r d θ . R f ( ϕ ( r , θ ) ) ω ϕ r , ϕ θ d r d θ . int_(R)f(phi(r,theta))omega((del phi)/(del r),(del phi)/(del theta))drd theta.\int_{R} f(\phi(r, \theta)) \omega\left(\frac{\partial \phi}{\partial r}, \frac{\partial \phi}{\partial \theta}\right) d r d \theta .Rf(ϕ(r,θ))ω(ϕr,ϕθ)drdθ.
In particular, if we examine linear functions for ω ω omega\omegaω, we arrive at a differential form. The moral is that if we want to perform an integral over a region parameterized by as in the previous section, then
we need to multiply by a function which takes a vector and returns a number. If we want to integrate over something parameterized by 2 , then we need to multiply by a function which takes two vectors and returns a number. In general, an n n nnn-form is a linear function which takes n n nnn vectors and returns a real number. One integrates n n nnn forms over regions that can be parameterized by n n nnn. Their strength is that the value of such an integral does not depend on the choice of parameterization.

4

Forms

4.1 Coordinates for vectors

Before we begin to discuss functions of vectors, we first need to learn how to specify a vector. And before we can answer that, we must first learn where vectors live. In Figure 4.1 we see a curve, C C CCC, and a tangent line to that curve. The line can be thought of as the set of all tangent vectors at the point, p p ppp. We denote that line as T p C T p C T_(p)CT_{p} CTpC, the tangent space to C C CCC at the point p p ppp.
Fig. 4.1. T p C T p C T_(p)CT_{p} CTpC is the set of all vectors tangents to C C CCC at p p ppp.
What if C C CCC is actually a straight line? Will T p C T p C T_(p)CT_{p} CTpC be the same line? To answer this, let's instead think about the real number line, L = 1 L = 1 L=1L=1L=1. Suppose p p ppp is the point corresponding to the number 2 on L L LLL. We would like to understand T p L T p L T_(p)LT_{p} LTpL, the set of all vectors tangent to L L LLL at the point p p ppp. For example, where would you draw a vector of length three? Would you put its base at the origin on L L LLL ? Of course not. You
would put its base at the point p p ppp. This is really because the origin for T p L T p L T_(p)LT_{p} LTpL is different than the origin for L L LLL. We are thus thinking about L L LLL and T p L T p L T_(p)LT_{p} LTpL as two different lines, placed right on top of each other.
The key to understanding the difference between L L LLL and T p L T p L T_(p)LT_{p} LTpL is their coordinate systems. Let's pause here for a moment to look a little more closely. What are "coordinates" anyway? They are a way of assigning a number (or, more generally, a set of numbers) to a point in space. In other words, coordinates are functions which take points of a space and return (sets of) numbers. When we say that the x x xxx-coordinate of p p ppp in 2 2 ^(2){ }^{2}2 is 5 , we really mean that we have a function, x : R 2 x : R 2 x:R^(2)rarrx: \mathbb{R}^{2} \rightarrowx:R2, such that x ( p ) = 5 x ( p ) = 5 x(p)=5x(p)=5x(p)=5.
Of course we need two numbers to specify a point in a plane, which means that we have two coordinate functions. Suppose we denote the plane by P P PPP and x : P R x : P R x:P rarrRx: P \rightarrow \mathbb{R}x:PR and y : P a r y : P a r y:P rarrary: P \rightarrow \mathbb{a r}y:Par coordinate functions. Then, saying that the coordinates of a point, p , p p_(", ")p_{\text {, }}p are ( 2 , 3 ) ( 2 , 3 ) (2,3)(2,3)(2,3) is the same thing as saying that x ( p ) = 2 x ( p ) = 2 x(p)=2x(p)=2x(p)=2 and y ( p ) = 3 y ( p ) = 3 y(p)=3y(p)=3y(p)=3. In other words, the coordinates of p p ppp are ( x ( p ) , y ( p ) ) ( x ( p ) , y ( p ) ) (x(p),y(p))(x(p), y(p))(x(p),y(p)).
So what do we use for coordinates in the tangent space? Well, first we need a basis for the tangent space of P P PPP at p p ppp. In other words, we need to pick two vectors which we can use to give the relative positions of all other points. Note that if the coordinates of p p ppp are ( x x xxx, y) then d ( x + t , y ) d t = 1 , 0 d ( x + t , y ) d t = 1 , 0 (d(x+t,y))/(dt)=(:1,0:)\frac{d(x+t, y)}{d t}=\langle 1,0\rangled(x+t,y)dt=1,0 and d ( x , y + t ) d t = 0 , 1 d ( x , y + t ) d t = 0 , 1 (d(x,y+t))/(dt)=(:0,1:)\frac{d(x, y+t)}{d t}=\langle 0,1\rangled(x,y+t)dt=0,1. We have switched to the notation " , , (:*,*:)\langle\cdot, \cdot\rangle, " to indicate that we are not talking about points of P P PPP anymore, but rather vectors in T p P T p P T_(p)PT_{p} PTpP. We take these two vectors to be a basis for T p P T p P T_(p)PT_{p} PTpP. In other words, any point of T p P T p P T_(p)PT_{p} PTpP can be written as d x 0 , 1 + d y 1 , 0 d x 0 , 1 + d y 1 , 0 dx(:0,1:)+dy(:1,0:)d x\langle 0,1\rangle+d y\langle 1,0\rangledx0,1+dy1,0, where d x , d y R d x , d y R dx,dy inRd x, d y \in \mathbb{R}dx,dyR. Hence, " d x d x dxd xdx " and " d y d y dyd ydy " are coordinate functions for T p P T p P T_(p)PT_{p} PTpP. Saying that the coordinates of a vector V V VVV in T p P T p P T_(p)PT_{p} PTpP are 2 , 3 2 , 3 (:2,3:)\langle 2,3\rangle2,3, for example, is the same thing as saying that d x ( V ) = 2 d x ( V ) = 2 dx(V)=2d x(V)=2dx(V)=2 and d y ( V ) = 3 d y ( V ) = 3 dy(V)=3d y(V)=3dy(V)=3. In general, we may refer to the coordinates of an arbitrary vector in T p P T p P T_(p)PT_{p} PTpP as d x , d y d x , d y (:dx,dy:)\langle d x, d y\rangledx,dy, just as we may refer to the coordinates of an arbitrary point in P P PPP as ( x , y ) ( x , y ) (x,y)(x, y)(x,y).
It will be helpful in the future to be able to distinguish between the vector 2 , 3 2 , 3 (:2,3:)\langle 2,3\rangle2,3 in T p P T p P T_(p)PT_{p} PTpP and the vector 2 , 3 2 , 3 (:2,3:)\langle 2,3\rangle2,3 in T q P T q P T_(q)PT_{q} PTqP, where p q p q p!=qp \neq qpq.
We will do this by writing 2 , 3 p 2 , 3 p (:2,3:)_(p)\langle 2,3\rangle_{p}2,3p for the former and 2 , 3 q 2 , 3 q (:2,3:)_(q)\langle 2,3\rangle_{q}2,3q for the latter.
Let's pause for a moment to address something that may have been bothering you since your first term of calculus. Let's look at the tangent line to the graph of y = x 2 y = x 2 y=x^(2)y=x^{2}y=x2 at the point ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1). We are no longer thinking of this tangent line as lying in the same plane that the graph does. Rather, it lies in T ( 1 , 1 ) 2 T ( 1 , 1 ) 2 T_((1,1))^(2)T_{(1,1)}{ }^{2}T(1,1)2. The horizontal axis for T ( 1 , 1 ) 2 T ( 1 , 1 ) 2 T_((1,1))^(2)T_{(1,1)}{ }^{2}T(1,1)2 is the " d x d x dx^(')d x^{\prime}dx axis and the vertical axis is the " d y d y dy^(')d y^{\prime}dy axis (see Fig. 4.2). Hence, we can write the equation of the tangent line as d y = d y = dy=d y=dy= 2 d x 2 d x 2dx2 d x2dx. We can rewrite this as d y d x = 2 d y d x = 2 (dy)/(dx)=2\frac{d y}{d x}=2dydx=2. Look familiar? This is one explanation of why we use the notation d y d x d y d x (dy)/(dx)\frac{d y}{d x}dydx in calculus to denote the derivative.

4.1.

  1. Draw a vector with d x = 1 , d y = 2 d x = 1 , d y = 2 dx=1,dy=2d x=1, d y=2dx=1,dy=2 in the tangent space T ( 1 , 1 ) T ( 1 , 1 ) T_((1,-1))T_{(1,-1)}T(1,1) 2.
  2. Draw 3 , 1 ( 0 , 1 ) 3 , 1 ( 0 , 1 ) (:-3,1:)_((0,1))\langle-3,1\rangle_{(0,1)}3,1(0,1).
Fig. 4.2. The line, l l lll, lies in T ( 1 , 1 ) 2 T ( 1 , 1 ) 2 T_((1,1))^(2)T_{(1,1)}{ }^{2}T(1,1)2. Its equation is d y = 2 d x d y = 2 d x dy=2dxd y=2 d xdy=2dx.

4.2 1-forms

Recall from the previous chapter, that a 1 -form is a linear function which acts on vectors and returns numbers. For the moment let's just look at 1 -forms on T p 2 T p 2 T_(p)^(2)T_{p}{ }^{2}Tp2 for some fixed point, p p ppp. Recall that a linear function, ω ω omega\omegaω, is just one whose graph is a plane through the origin. Hence, we want to write down an equation of a plane though the origin in T p 2 × T p 2 × T_(p)^(2)xxT_{p}{ }^{2} \timesTp2×, where one axis is labelled d x d x dxd xdx, another d y d y dyd ydy and the third, ω ω omega\omegaω (see Fig. 4.3). This is easy: ω = a d x + b d y ω = a d x + b d y omega=adx+bdy\omega=a d x+b d yω=adx+bdy. Hence, to specify a 1 -form on T p 2 T p 2 T_(p)^(2)T_{p}{ }^{2}Tp2 we only need to know two numbers: a a aaa and b b bbb.
Here is a quick example. Suppose ω ( d x , d y ) = 2 d x + 3 d y ω ( d x , d y ) = 2 d x + 3 d y omega((:dx,dy:))=2dx+3dy\omega(\langle d x, d y\rangle)=2 d x+3 d yω(dx,dy)=2dx+3dy, then
ω ( { 1 , 2 ) = 2 1 + 3 2 = 4 . ω ( { 1 , 2 ) = 2 1 + 3 2 = 4 . omega({-1,2:))=2*-1+3*2=4.\omega(\{-1,2\rangle)=2 \cdot-1+3 \cdot 2=4 .ω({1,2)=21+32=4.
The alert reader may see something familiar here: the dot product. That is, ω ( 1 , 2 ) = 2 , 3 1 , 2 ω ( 1 , 2 ) = 2 , 3 1 , 2 omega((:-1,2:))=(:2,3:)*(:-1,2:)\omega(\langle-1,2\rangle)=\langle 2,3\rangle \cdot\langle-1,2\rangleω(1,2)=2,31,2. Recall the geometric interpretation of the dot product; you project 1 , 2 1 , 2 (:-1,2:)\langle-1,2\rangle1,2 onto 2 , 3 2 , 3 (:2,3:)\langle 2,3\rangle2,3 and then multiply by | 2 , 3 | = 13 | 2 , 3 | = 13 |(:2,3:)|=sqrt()13|\langle 2,3\rangle|=\sqrt{ } 13|2,3|=13. In other words:
Evaluating a 1 -form on a vector is the same as projecting onto some line and then multiplying by some constant.
Fig. 4.3. The graph of ω ω omega\omegaω is a plane though the origin.
In fact, we can even interpret the act of multiplying by a constant geometrically. Suppose ω ω omega\omegaω is given by a d x + b d y a d x + b d y adx+bdya d x+b d yadx+bdy. Then the value of ω ( V 1 ) ω V 1 omega(V_(1))\omega\left(V_{1}\right)ω(V1) is the length of the projection of V 1 V 1 V_(1)V_{1}V1 onto the line, I I III, where a , b | a , b | 2 a , b | a , b | 2 ((:a,b:))/(|(:a,b:)|^(2))\frac{\langle a, b\rangle}{|\langle a, b\rangle|^{2}}a,b|a,b|2 is a basis vector for l l lll.
This interpretation has a huge advantage... it is coordinate free. Recall from the previous section that we can think of the plane, P P PPP, as existing independent of our choice of coordinates. We only pick coordinates so that we can communicate to someone else the location of a point. Forms are similar. They are objects that exist independently of our choice of coordinates. This is one key as to why they are so useful outside of mathematics.
There is still another geometric interpretation of 1-forms. Let's first look at the simple example ω ( d x , d y ) = d x ω ( d x , d y ) = d x omega((:dx,dy:))=dx\omega(\langle d x, d y\rangle)=d xω(dx,dy)=dx. This 1 -form simply returns the first coordinate of whatever vector you feed into it. This is also a projection; it's the projection of the input vector onto the d x d x dxd xdx-axis. This immediately gives us a new interpretation of the action of a general 1-form, ω = a d x + b d y ω = a d x + b d y omega=adx+bdy\omega=a d x+b d yω=adx+bdy.
Evaluating a 1 -form on a vector is the same as projecting onto each coordinate axis, scaling each by some constant and adding the results.
Although this interpretation is more cumbersome, it is the one that will generalize better when we get to n n nnn-forms.
Let's move on now to 1 -forms in n n nnn dimensions. If p n p n p in np \in npn, then we can write p p ppp in coordinates as ( x 1 , x 2 , , x n ) x 1 , x 2 , , x n (x_(1),x_(2),dots,x_(n))\left(x_{1}, x_{2}, \ldots, x_{n}\right)(x1,x2,,xn). The coordinates for a vector in T p n T p n T_(p)^(n)T_{p}{ }^{n}Tpn are d x 1 , d x 2 , , d x n d x 1 , d x 2 , , d x n (:dx_(1),dx_(2),dots,dx_(n):)\left\langle d x_{1}, d x_{2}, \ldots, d x_{n}\right\rangledx1,dx2,,dxn. A 1-form is a linear function, ω ω omega\omegaω, whose graph (in T p n × T p n × T_(p)^(n)xxT_{p}{ }^{n} \timesTpn× ) is a plane through the origin. Hence, we can write it as ω = a 1 d x 1 + a 2 d x 2 + + a n d x n ω = a 1 d x 1 + a 2 d x 2 + + a n d x n omega=a_(1)dx_(1)+a_(2)dx_(2)+dots+a_(n)dx_(n)\omega=a_{1} d x_{1}+a_{2} d x_{2}+\ldots+a_{n} d x_{n}ω=a1dx1+a2dx2++andxn. Again, this can be thought of as either projecting onto the vector a 1 , a 2 , , a n a 1 , a 2 , , a n (:a_(1),a_(2),dots,a_(n):)\left\langle a_{1}, a_{2}, \ldots, a_{n}\right\ranglea1,a2,,an and then multiplying by | a 1 , a 2 , , a n | a 1 , a 2 , , a n |(:a_(1),a_(2),dots,a_(n):)|\left|\left\langle a_{1}, a_{2}, \ldots, a_{n}\right\rangle\right||a1,a2,,an| or as projecting onto each coordinate axis, multiplying by a j a j a_(j)a_{j}aj, and then adding.
4.2. Let ω ( d x , d y ) = d x + 4 d y ω ( d x , d y ) = d x + 4 d y omega((:dx,dy:))=-dx+4dy\omega(\langle d x, d y\rangle)=-d x+4 d yω(dx,dy)=dx+4dy.
  1. Compute ω ( 1 , 0 ) , ω ( 0 , 1 ) ω ( 1 , 0 ) , ω ( 0 , 1 ) omega((:1,0:)),omega((:0,1:))\omega(\langle 1,0\rangle), \omega(\langle 0,1\rangle)ω(1,0),ω(0,1) and ω ( 2 , 3 ) ω ( 2 , 3 ) omega(2,3:))\omega(2,3\rangle)ω(2,3).
  2. What line does ω ω omega\omegaω project vectors onto?
    4.3. Find a 1 -form which computes the length of the projection of a vector onto the indicated line, multiplied by the indicated constant c c ccc.
  3. The d x d x dxd xdx-axis, c = 3 c = 3 c=3c=3c=3.
  4. The d y d y dyd ydy-axis, c = 1 / 2 c = 1 / 2 c=1//2c=1 / 2c=1/2.
  5. Find a 1 -form that does both of the two preceding operations and adds the result.
  6. The line d y = 3 / 4 d x , c = 10 d y = 3 / 4 d x , c = 10 dy=3//4dx,c=10d y=3 / 4 d x, c=10dy=3/4dx,c=10.
    4.4. If ω ω omega\omegaω is a 1 -form show
  7. ω ( V 1 + V 2 ) = ω ( V 1 ) + ω ( V 2 ) ω V 1 + V 2 = ω V 1 + ω V 2 omega(V_(1)+V_(2))=omega(V_(1))+omega(V_(2))\omega\left(V_{1}+V_{2}\right)=\omega\left(V_{1}\right)+\omega\left(V_{2}\right)ω(V1+V2)=ω(V1)+ω(V2), for any vectors V 1 V 1 V_(1)V_{1}V1 and V 2 V 2 V_(2)V_{2}V2.
  8. ω ( c V ) = c ω ( V ) ω ( c V ) = c ω ( V ) omega(cV)=c omega(V)\omega(c V)=c \omega(V)ω(cV)=cω(V), for any vector V V VVV and constant c c ccc.

4.3 Multiplying 1-forms

In this section we would like to explore a method of multiplying 1forms. You may think, "What is the big deal? If ω ω omega\omegaω and v v vvv are 1 -forms can't we just define ω v ( V ) = ω ( V ) v ( V ) ω v ( V ) = ω ( V ) v ( V ) omega*v(V)=omega(V)*v(V)\omega \cdot v(V)=\omega(V) \cdot v(V)ωv(V)=ω(V)v(V) ?" Well, of course we can, but then ω v ω v omega*v\omega \cdot vωv is not a linear function, so we have left the world of forms.
The trick is to define the product of ω ω omega\omegaω and v v vvv to be a 2 -form. So as not to confuse this with the product just mentioned, we will use the symbol " Λ Λ Lambda\LambdaΛ " (pronounced "wedge") to denote multiplication. So how can we possibly define ω ω omega^^vv\omega \wedge \veeω to be a 2-form? We must define how it acts on a pair of vectors, ( V 1 , V 2 ) V 1 , V 2 (V_(1),V_(2))\left(V_{1}, V_{2}\right)(V1,V2).
Note first that there are four ways to combine all the ingredients:
ω ( V 1 ) , v ( V 1 ) , ω ( V 2 ) , v ( V 2 ) . ω V 1 , v V 1 , ω V 2 , v V 2 . omega(V_(1)),v(V_(1)),omega(V_(2)),v(V_(2)).\omega\left(V_{1}\right), v\left(V_{1}\right), \omega\left(V_{2}\right), v\left(V_{2}\right) .ω(V1),v(V1),ω(V2),v(V2).
The first two of these are associated with V 1 V 1 V_(1)V_{1}V1 and the second two with V 2 V 2 V_(2)V_{2}V2. In other words, ω ω omega\omegaω and v v vvv together give a way of taking each vector and returning a pair of numbers. And how do we visualize pairs of numbers? In the plane, of course! Let's define a new plane with one axis as the ω ω omega\omegaω-axis and the other as the v v vvv-axis. So, the coordinates of V 1 V 1 V_(1)V_{1}V1 in this plane are [ ω ( V 1 ) , v ( V 1 ) ] ω V 1 , v V 1 [omega(V_(1)),v(V_(1))]\left[\omega\left(V_{1}\right), \mathrm{v}\left(V_{1}\right)\right][ω(V1),v(V1)] and the coordinates of V 2 V 2 V_(2)V_{2}V2 are [ ω ( V 2 ) , v ( V 2 ) ] ω V 2 , v V 2 [omega(V_(2)),v(V_(2))]\left[\omega\left(V_{2}\right), \mathrm{v}\left(V_{2}\right)\right][ω(V2),v(V2)]. Note that we have switched to the notation " [ , ] [ , ] [:',*][\because, \cdot][,] " to indicate that we are describing points in a new plane. This may seem a little confusing at first. Just keep in mind that when we write something like ( 1 , 2 ) ( 1 , 2 ) (1,2)(1,2)(1,2) we are describing the location of a point in the x y x y xyx yxy-plane, whereas 1 , 2 1 , 2 (:1,2:)\langle 1,2\rangle1,2 describes a vector in the d x d y d x d y dxdyd x d ydxdy-plane and [ 1 , 2 ] [ 1 , 2 ] [1,2][1,2][1,2] is a vector in the ω V ω V omega V\omega VωV-plane.
Let's not forget our goal now. We wanted to use ω ω omega\omegaω and v v vvv to take the pair of vectors, ( V 1 , V 2 ) V 1 , V 2 (V_(1),V_(2))\left(V_{1}, V_{2}\right)(V1,V2), and return a number. So far all we have done is to take this pair of vectors and return another pair of vectors. But do we know of a way to take these vectors and get a number? Actually, we know several, but the most useful one turns out to be the area of the parallelogram that the vectors span. This is precisely what we define to be the value of ω v ( V 1 , V 2 ) ω v V 1 , V 2 omega^^v(V_(1),V_(2))\omega \wedge v\left(V_{1}, V_{2}\right)ωv(V1,V2) (see Fig. 4.4).
Fig. 4.4. The product of ω ω omega\omegaω and v v vvv.
Example 14. Let ω = 2 d x 3 d y + d z ω = 2 d x 3 d y + d z omega=2dx-3dy+dz\omega=2 d x-3 d y+d zω=2dx3dy+dz and v = d x + 2 d y d z v = d x + 2 d y d z v=dx+2dy-dz\mathrm{v}=d x+2 d y-d zv=dx+2dydz be two 1-forms on T p 3 T p 3 T_(p)^(3)T_{p}{ }^{3}Tp3 for some fixed p } 3 p 3 p in}^(3)\left.p \in\right\}^{3}p}3. Let's evaluate ω ω omega^^vv\omega \wedge \veeω on the pair of vectors, ( 1 , 3 , 1 , 2 , 1 , 3 ) ( 1 , 3 , 1 , 2 , 1 , 3 ) ((:1,3,1:),(:2,-1,3:))(\langle 1,3,1\rangle,\langle 2,-1,3\rangle)(1,3,1,2,1,3). First we compute the [ ω , v ] [ ω , v ] [omega,v][\omega, v][ω,v] coordinates of the vector 1 , 3 , 1 1 , 3 , 1 (:1,3,1:)\langle 1,3,1\rangle1,3,1.
[ ω ( 1 , 3 , 1 ) , v ( 1 , 3 , 1 ) ] = [ 2 1 3 3 + 1 1 , 1 1 + 2 3 1 1 ] = [ 6 , 6 ] . [ ω ( 1 , 3 , 1 ) , v ( 1 , 3 , 1 ) ] = [ 2 1 3 3 + 1 1 , 1 1 + 2 3 1 1 ] = [ 6 , 6 ] . {:[[omega((:1","3","1:))","v((:1","3","1:))]=[2*1-3*3+1*1","1*1+2*3-1*1]],[=[-6","6].]:}\begin{aligned} {[\omega(\langle 1,3,1\rangle), v(\langle 1,3,1\rangle)] } & =[2 \cdot 1-3 \cdot 3+1 \cdot 1,1 \cdot 1+2 \cdot 3-1 \cdot 1] \\ & =[-6,6] . \end{aligned}[ω(1,3,1),v(1,3,1)]=[2133+11,11+2311]=[6,6].
Similarly, we compute [ ω ( 2 , 1 , 3 ) , v ( 2 , 1 , 3 ) ] = [ 10 , 3 ] [ ω ( 2 , 1 , 3 ) , v ( 2 , 1 , 3 ) ] = [ 10 , 3 ] [omega((:2,-1,3:)),v(2,-1,3:))]=[10,-3][\omega(\langle 2,-1,3\rangle), \mathrm{v}(2,-1,3\rangle)]=[10,-3][ω(2,1,3),v(2,1,3)]=[10,3]. Finally, the area of the parallelogram spanned by [ 6 , 6 ] [ 6 , 6 ] [-6,6][-6,6][6,6] and [ 10 , 3 ] [ 10 , 3 ] [10,-3][10,-3][10,3] is
| 6 10 6 3 | = 18 60 = 42 . 6      10 6      3 = 18 60 = 42 . |[-6,10],[6,-3]|=18-60=-42.\left|\begin{array}{rr} -6 & 10 \\ 6 & -3 \end{array}\right|=18-60=-42 .|61063|=1860=42.
Should we have taken the absolute value? Not if we want to define a linear operator. The result of ω v ω v omega^^v\omega \wedge vωv is not just an area, it is a signed area; it can either be positive or negative. We will see a geometric interpretation of this soon. For now, we define:
ω v ( V 1 , V 2 ) = | ω ( V 1 ) ω ( V 2 ) ν ( V 1 ) v ( V 2 ) | . ω v V 1 , V 2 = ω V 1      ω V 2 ν V 1      v V 2 . omega^^v(V_(1),V_(2))=|[omega(V_(1)),omega(V_(2))],[nu(V_(1)),v(V_(2))]|.\omega \wedge v\left(V_{1}, V_{2}\right)=\left|\begin{array}{ll} \omega\left(V_{1}\right) & \omega\left(V_{2}\right) \\ \nu\left(V_{1}\right) & v\left(V_{2}\right) \end{array}\right| .ωv(V1,V2)=|ω(V1)ω(V2)ν(V1)v(V2)|.
4.5. Let ω ω omega\omegaω and v v vvv be the following 1 -forms:
ω ( ( d x , d y ) = 2 d x 3 d y v ( d x , d y ) = d x + d y ω ( ( d x , d y ) = 2 d x 3 d y v ( d x , d y ) = d x + d y {:[omega((dx","dy:))=2dx-3dy],[v((:dx","dy:))=dx+dy]:}\begin{gathered} \omega((d x, d y\rangle)=2 d x-3 d y \\ v(\langle d x, d y\rangle)=d x+d y \end{gathered}ω((dx,dy)=2dx3dyv(dx,dy)=dx+dy
  1. Let V 1 = 1 , 2 V 1 = 1 , 2 V_(1)=(:-1,2:)V_{1}=\langle-1,2\rangleV1=1,2 and V 2 = 1 , 1 V 2 = 1 , 1 V_(2)=(:1,1:)V_{2}=\langle 1,1\rangleV2=1,1. Compute ω ( V 1 ) , v ( V 1 ) , ω ( V 2 ω V 1 , v V 1 , ω V 2 omega(V_(1)),v(V_(1)),omega(V_(2):}\omega\left(V_{1}\right), v\left(V_{1}\right), \omega\left(V_{2}\right.ω(V1),v(V1),ω(V2 ) and V ( V 2 ) V V 2 V(V_(2))\mathrm{V}\left(V_{2}\right)V(V2).
  2. Use your answers to the previous question to compute ω ω omega^^vv\omega \wedge \veeω ( V 1 , V 2 ) V 1 , V 2 (V_(1),V_(2))\left(V_{1}, V_{2}\right)(V1,V2).
  3. Find a constant c c ccc such that ω v = c d x d y ω v = c d x d y omega^^v=cdx^^dy\omega \wedge v=c d x \wedge d yωv=cdxdy.
    4.6. ω v ( V 1 , V 2 ) = ω v ( V 2 , V 1 ) ( ω v ω v V 1 , V 2 = ω v V 2 , V 1 ( ω v omega^^v(V_(1),V_(2))=-omega^^v(V_(2),V_(1))(omega^^v\omega \wedge v\left(V_{1}, V_{2}\right)=-\omega \wedge v\left(V_{2}, V_{1}\right)(\omega \wedge vωv(V1,V2)=ωv(V2,V1)(ωv is skew-symmetric).
    4.7. ω v ( V , V ) = 0 ω v ( V , V ) = 0 omega^^v(V,V)=0\omega \wedge v(V, V)=0ωv(V,V)=0. (This follows immediately from the previous exercise. It should also be clear from the geometric interpretation.)
    4.8. ω v ( V 1 + V 2 , V 3 ) = ω v ( V 1 , V 3 ) + ω v ( V 2 , V 3 ) ω v V 1 + V 2 , V 3 = ω v V 1 , V 3 + ω v V 2 , V 3 omega^^v(V_(1)+V_(2),V_(3))=omega^^v(V_(1),V_(3))+omega^^v(V_(2),V_(3))\omega \wedge v\left(V_{1}+V_{2}, V_{3}\right)=\omega \wedge v\left(V_{1}, V_{3}\right)+\omega \wedge v\left(V_{2}, V_{3}\right)ωv(V1+V2,V3)=ωv(V1,V3)+ωv(V2,V3) and ω ω omega^^\omega \wedgeω v ( c V 1 , V 2 ) = ω v ( V 1 , c V 2 ) = c ω v ( V 1 , V 2 ) v c V 1 , V 2 = ω v V 1 , c V 2 = c ω v V 1 , V 2 v(cV_(1),V_(2))=omega^^v(V_(1),cV_(2))=c omega^^v(V_(1),V_(2))v\left(c V_{1}, V_{2}\right)=\omega \wedge v\left(V_{1}, c V_{2}\right)=c \omega \wedge v\left(V_{1}, V_{2}\right)v(cV1,V2)=ωv(V1,cV2)=cωv(V1,V2), where c c ccc is any real number ( ω ω omega^^vv\omega \wedge \veeω is bilinear).
    4.9. ω ( V 1 , V 2 ) = ω ( V 1 , V 2 ) ω V 1 , V 2 = ω V 1 , V 2 omega^^vv(V_(1),V_(2))=-vv^^omega(V_(1),V_(2))\omega \wedge \vee\left(V_{1}, V_{2}\right)=-\vee \wedge \omega\left(V_{1}, V_{2}\right)ω(V1,V2)=ω(V1,V2).
It is interesting to compare Problems 4.6 and 4.9. Problem 4.6 says that the 2 -form, ω v ω v omega^^v\omega \wedge vωv, is a skew-symmetric operator on pairs of vectors. Problem 4.9 says that ^^\wedge can be thought of as a skewsymmetric operator on 1 -forms.
4.10. ω ω ( V 1 , V 2 ) = 0 ω ω V 1 , V 2 = 0 omega^^omega(V_(1),V_(2))=0\omega \wedge \omega\left(V_{1}, V_{2}\right)=0ωω(V1,V2)=0.
4.11. ( ω + ) Ψ = ω Ψ + Ψ ( ( ω + ) Ψ = ω Ψ + Ψ ( (omega+vv)^^Psi=omega^^Psi+vv^^Psi(^^(\omega+\vee) \wedge \Psi=\omega \wedge \Psi+\vee \wedge \Psi(\wedge(ω+)Ψ=ωΨ+Ψ( is distributive ) ) ))).
There is another way to interpret the action of ω v ω v omega^^v\omega \wedge vωv which is much more geometric. First let ω = a d x + b d y ω = a d x + b d y omega=adx+bdy\omega=a d x+b d yω=adx+bdy be a 1-form on T p T p T_(p)T_{p}Tp 2 2 ^(2){ }^{2}2. Then we let ω ω (:omega:)\langle\omega\rangleω be the vector a , b a , b (:a,b:)\langle a, b\ranglea,b.
4.12. Let ω ω omega\omegaω and v v vvv be 1 -forms on T p T p T_(p)T_{p}Tp. Show that ω v ( V 1 , V 2 ) ω v V 1 , V 2 omega^^v(V_(1),V_(2))\omega \wedge v\left(V_{1}, V_{2}\right)ωv(V1,V2) is the area of the parallelogram spanned by V 1 V 1 V_(1)V_{1}V1 and V 2 V 2 V_(2)V_{2}V2, times the area of the parallelogram spanned by ω ω (:omega:)\langle\omega\rangleω and v v (:v:)\langle\mathrm{v}\ranglev.
4.13. Use the previous problem to show that if ω ω omega\omegaω and v v vvv are 1 -forms on 2 2 ^(2){ }^{2}2 such that ω v = 0 ω v = 0 omega^^v=0\omega \wedge v=0ωv=0 then there is a constant c c ccc such that ω = ω = omega=\omega=ω= cV.
There is also a more geometric way to think about ω v ω v omega^^v\omega \wedge vωv if ω ω omega\omegaω and V are 1 -forms on T ρ 3 T ρ 3 T_(rho)^(3)T_{\rho}{ }^{3}Tρ3, although it will take us some time to develop the idea. Suppose ω = a d x + b d y + c d z ω = a d x + b d y + c d z omega=adx+bdy+cdz\omega=a d x+b d y+c d zω=adx+bdy+cdz. Then we will denote the vector a , b , c a , b , c (:a,b,c:)\langle a, b, c\ranglea,b,c as ω ω (:omega:)\langle\omega\rangleω. From the previous section, we know that if V V VVV is any vector, then ω ( V ) = ω v ω ( V ) = ω v omega(V)=(:omega:)*v\omega(V)=\langle\omega\rangle \cdot vω(V)=ωv, and that this is just the projection of V V VVV onto the line containing ω ω (:omega:)\langle\omega\rangleω, times | ω | | ω | |(:omega:)||\langle\omega\rangle||ω|.
Now suppose v is some other 1-form. Choose a scalar x x xxx so that v v v v v_(v)\mathrm{v}_{\mathrm{v}}vv x ω x ω -x omega:)-x \omega\ranglexω is perpendicular to ω ω (:omega:)\langle\omega\rangleω. Let v ω = v x ω v ω = v x ω v_(omega)=v-x omegav_{\omega}=v-x \omegavω=vxω. Note that ω v ω ω v ω omega^^v_(omega)\omega \wedge v_{\omega}ωvω = ω ( v x ω ) = ω v x ω ω = ω v = ω ( v x ω ) = ω v x ω ω = ω v =omega^^(v-x omega)=omega^^v-x omega^^omega=omega^^v=\omega \wedge(v-x \omega)=\omega \wedge v-x \omega \wedge \omega=\omega \wedge v=ω(vxω)=ωvxωω=ωv. Hence, any geometric interpretation we find for the action of ω v ω ω v ω omega^^v_(omega)\omega \wedge v_{\omega}ωvω is also a geometric interpretation of the action of ω v ω v omega^^v\omega \wedge vωv.
Finally, we let ω ¯ = ω | ω | ω ¯ = ω | ω | ^( bar(omega))=(omega)/(|(:omega:)|){ }^{\bar{\omega}}=\frac{\omega}{|\langle\omega\rangle|}ω¯=ω|ω| and ω ¯ ω ¯ = v ω | ω ω ω ¯ ω ¯ = v ω ω ω ^( bar(omega)) bar(omega)=(v_(omega))/(|(:omega_(omega):):))^{\bar{\omega}} \bar{\omega}=\frac{v_{\omega}}{\left|\left\langle\omega_{\omega}\right\rangle\right\rangle}ω¯ω¯=vω|ωω. Note that these are 1forms such that ω ω (:omega:)\langle\omega\rangleω and v ω v ω (:v_(omega):)\left\langle v_{\omega}\right\ranglevω are perpendicular unit vectors. We will now present a geometric interpretation of the action of ω v ω ω v ω omega^^v_(omega)\omega \wedge v_{\omega}ωvω on a pair of vectors, ( V 1 , V 2 ) V 1 , V 2 (V_(1),V_(2))\left(V_{1}, V_{2}\right)(V1,V2).
First, note that since ω ω (:omega:)\langle\omega\rangleω is a unit vector then ω ( V 1 ) ω V 1 omega(V_(1))\omega\left(V_{1}\right)ω(V1) is just the projection of V 1 V 1 V_(1)V_{1}V1 onto the line containing ω ω (:omega:)\langle\omega\rangleω. Similarly, v ω ( V 1 ) v ω V 1 v_(omega)(V_(1))v_{\omega}\left(V_{1}\right)vω(V1) is
given by projecting V 1 V 1 V_(1)V_{1}V1 onto the line containing v ω v ω (:v omega:)\langle v \omega\ranglevω. As ω ω (:omega:)\langle\omega\rangleω and v v (:v\langle vv ω ω omega\omegaω ) are perpendicular, we can think of the quantity as the area of parallelogram spanned by V 1 V 1 V_(1)V_{1}V1 and V 2 V 2 V_(2)V_{2}V2, projected onto the plane containing the vectors ω ω (:omega:)\langle\omega\rangleω and v ω v ω (:v_(omega):)\left\langle v_{\omega}\right\ranglevω. This is the same plane as the one which contains the vectors ω ω (:omega:)\langle\omega\rangleω and v v (:v:)\langle\mathrm{v}\ranglev.
ω ¯ v ω ( V 1 , V 2 ) = | ω ¯ ( V 1 ) v ω ( V 2 ) ω ¯ 1 ν ω ( V 2 ) | ω ¯ v ω ¯ V 1 , V 2 = ω ¯ V 1      v ω ¯ V 2 ω ¯ 1      ν ω ¯ V 2 bar(omega)^^ bar(v_(omega))(V_(1),V_(2))=|[ bar(omega)(V_(1)), bar(v_(omega))(V_(2))],[ bar(omega)_(1), bar(nu_(omega))(V_(2))]|\bar{\omega} \wedge \overline{v_{\omega}}\left(V_{1}, V_{2}\right)=\left|\begin{array}{ll} \bar{\omega}\left(V_{1}\right) & \overline{v_{\omega}}\left(V_{2}\right) \\ \bar{\omega}_{1} & \overline{\nu_{\omega}}\left(V_{2}\right) \end{array}\right|ω¯vω(V1,V2)=|ω¯(V1)vω(V2)ω¯1νω(V2)|
Now observe the following:
ω ¯ ω ω = ω | ( ω ) | v ω | v ω | = 1 | ω ) | v ω ) | ω v ω . ω ¯ ω ω ¯ = ω | ( ω ) | v ω v ω = 1 ω ) v ω ω v ω . bar(omega)^^ bar(omega_(omega))=(omega)/(|(omega)|∣)^^(v_(omega))/(|(:v_(omega):)|∣)=(1)/(|(:omega)|||(:v_(omega))|)omega^^v_(omega).\bar{\omega} \wedge \overline{\omega_{\omega}}=\frac{\omega}{|(\omega)| \mid} \wedge \frac{v_{\omega}}{\left|\left\langle v_{\omega}\right\rangle\right| \mid}=\frac{1}{\left|\langle\omega) \|\left|\left\langle v_{\omega}\right)\right|\right.} \omega \wedge v_{\omega} .ω¯ωω=ω|(ω)|vω|vω|=1|ω)|vω)|ωvω.
Hence,
ω ν = ω v ω = ω v ω ) ω ¯ ω ω ω ν = ω v ω = ω v ω ω ¯ ω ω ¯ omega^^nu=omega^^v_(omega)=||(:omega:)||(:v_(omega))∣ bar(omega)^^ bar(omega_(omega))\omega \wedge \nu=\omega \wedge v_{\omega}=\|\langle\omega\rangle\|\left\langle v_{\omega}\right) \mid \bar{\omega} \wedge \overline{\omega_{\omega}}ων=ωvω=ωvω)ω¯ωω
Finally, note that since ω ω (:omega:)\langle\omega\rangleω and v ω v ω (:v_(omega):)\left\langle v_{\omega}\right\ranglevω are perpendicular, the quantity | ω | | v ω | | ω | v ω |(:omega:)||(:v_(omega):)||\langle\omega\rangle|\left|\left\langle v_{\omega}\right\rangle\right||ω||vω| is just the area of the rectangle spanned by these two vectors. Furthermore, the parallelogram spanned by the vectors ω ω (:omega:)\langle\omega\rangleω and v v (:v:)\langle\mathrm{v}\ranglev is obtained from this rectangle by skewing. Hence, they have the same area. We conclude
Evaluating ω v ω v omega^^v\omega \wedge vωv on the pair of vectors ( V 1 , V 2 ) V 1 , V 2 (V_(1),V_(2))\left(V_{1}, V_{2}\right)(V1,V2) gives the area of parallelogram spanned by V 1 V 1 V_(1)V_{1}V1 and V 2 V 2 V_(2)V_{2}V2 projected onto the plane containing the vectors ω ω (:omega:)\langle\omega\rangleω and v v (:v:)\langle\mathrm{v}\ranglev, and multiplied by the area of the parallelogram spanned by ω ω (:omega:)\langle\omega\rangleω and v v (:v:)\langle\mathrm{v}\ranglev.
CAUTION: While every 1 -form can be thought of as projected length not every 2 -form can be thought of as projected area. The only 2 -forms for which this interpretation is valid are those that are the product of 1 -forms. See Problem 4.18.
Let's pause for a moment to look at a particularly simple 2 -form on T p 3 , d x d y T p 3 , d x d y T_(p)^(3),dx^^dyT_{p}{ }^{3}, d x \wedge d yTp3,dxdy. Suppose V 1 = a 1 , a 2 , a 3 V 1 = a 1 , a 2 , a 3 V_(1)=(:a_(1),a_(2),a_(3):)V_{1}=\left\langle a_{1}, a_{2}, a_{3}\right\rangleV1=a1,a2,a3 and V 2 = b 1 , b 2 , b 3 V 2 = b 1 , b 2 , b 3 V_(2)=(:b_(1),b_(2),b_(3):)V_{2}=\left\langle b_{1}, b_{2}, b_{3}\right\rangleV2=b1,b2,b3. Then
d x d y ( V 1 , V 2 ) = | a 1 b 1 a 2 b 2 | . d x d y V 1 , V 2 = a 1 b 1 a 2 b 2 . dx^^dy(V_(1),V_(2))=|[a_(1)b_(1)],[a_(2)]b_(2)|.d x \wedge d y\left(V_{1}, V_{2}\right)=\left|\begin{array}{l} a_{1} b_{1} \\ a_{2} \end{array} b_{2}\right| .dxdy(V1,V2)=|a1b1a2b2|.
This is precisely the (signed) area of the parallelogram spanned by V 1 V 1 V_(1)V_{1}V1 and V 2 V 2 V_(2)V_{2}V2 projected onto the d x d y d x d y dxdyd x d ydxdy-plane.
4.14. ω ( a 1 , a 2 , a 3 , b 1 , b 2 , b 3 ) = c 1 d x d y + c 2 d x d z + ω a 1 , a 2 , a 3 , b 1 , b 2 , b 3 = c 1 d x d y + c 2 d x d z + omega^^vv((:a_(1),a_(2),a_(3):),(:b_(1),b_(2),b_(3):))=c_(1)dx^^dy+c_(2)dx^^dz+\omega \wedge \vee\left(\left\langle a_{1}, a_{2}, a_{3}\right\rangle,\left\langle b_{1}, b_{2}, b_{3}\right\rangle\right)=c_{1} d x \wedge d y+c_{2} d x \wedge d z+ω(a1,a2,a3,b1,b2,b3)=c1dxdy+c2dxdz+ c 3 d y d z c 3 d y d z c_(3)dy^^dzc_{3} d y \wedge d zc3dydz, for some real numbers c 1 , c 2 c 1 , c 2 c_(1),c_(2)c_{1}, c_{2}c1,c2 and c 3 c 3 c_(3)c_{3}c3.
The preceding comments, and this last exercise, give the following geometric interpretation of the action of a 2-form on the pair of vectors, ( V 1 , V 2 ) V 1 , V 2 (V_(1),V_(2))\left(V_{1}, V_{2}\right)(V1,V2) :
Every 2-form projects the parallelogram spanned by V 1 V 1 V_(1)V_{1}V1 and V 2 V 2 V_(2)V_{2}V2 onto each of the (2-dimensional) coordinate planes, computes the resulting (signed) areas, multiplies each by some constant, and adds the results.
This interpretation holds in all dimensions. Hence, to specify a 2 form we need to know as many constants as there are 2dimensional coordinate planes. For example, to give a 2 -form in 4 dimensional Euclidean space we need to specify six numbers:
c 1 d x d y + c 2 d x d z + c 3 d x d w + c 4 d y d z + c 5 d y d w + c 6 d z d w c 1 d x d y + c 2 d x d z + c 3 d x d w + c 4 d y d z + c 5 d y d w + c 6 d z d w c_(1)dx^^dy+c_(2)dx^^dz+c_(3)dx^^dw+c_(4)dy^^dz+c_(5)dy^^dw+c_(6)dz^^dwc_{1} d x \wedge d y+c_{2} d x \wedge d z+c_{3} d x \wedge d w+c_{4} d y \wedge d z+c_{5} d y \wedge d w+c_{6} d z \wedge d wc1dxdy+c2dxdz+c3dxdw+c4dydz+c5dydw+c6dzdw
The skeptic may argue here. Problem 4.14 only shows that a 2 form which is a product of 1 -forms can be thought of as a sum of projected, scaled areas. What about an arbitrary 2 -form? Well, to address this, we need to know what an arbitrary 2 -form is! Up until now we have not given a complete definition. Henceforth, we shall define a 2 -form to be a bilinear, skew-symmetric, real-valued function on T p n × T p n T p n × T p n T_(p)^(n)xxT_(p)^(n)T_{p}{ }^{n} \times T_{p}{ }^{n}Tpn×Tpn. That is a mouthful. This just means that it is an operator which eats pairs of vectors, spits out real numbers, and satisfies the conclusions of Problems 4.6 and 4.8. Since these are the only ingredients necessary to do Problem 4.14, our geometric interpretation is valid for all 2 -forms.
4.15. If ω ( d x , d y , d z ) = d x + 5 d y d z ω ( d x , d y , d z ) = d x + 5 d y d z omega((:dx,dy,dz:))=dx+5dy-dz\omega(\langle d x, d y, d z\rangle)=d x+5 d y-d zω(dx,dy,dz)=dx+5dydz, and v ( d x , d y , d z ) = 2 d x v ( d x , d y , d z ) = 2 d x v((:dx,dy,dz:))=2dx-\mathrm{v}(\langle d x, d y, d z\rangle)=2 d x-v(dx,dy,dz)=2dx d y + d z d y + d z dy+dzd y+d zdy+dz, compute
ω v ( ( 1 , 2 , 3 ) , { 1 , 4 , 2 ) ) . ω v ( ( 1 , 2 , 3 ) , { 1 , 4 , 2 ) ) . omega^^v((1,2,3),{-1,4,-2)).\omega \wedge v((1,2,3),\{-1,4,-2)) .ωv((1,2,3),{1,4,2)).
4.16. Let ω ( d x , d y , d z ) = d x + 5 d y d z ω ( d x , d y , d z ) = d x + 5 d y d z omega((:dx,dy,dz:))=dx+5dy-dz\omega(\langle d x, d y, d z\rangle)=d x+5 d y-d zω(dx,dy,dz)=dx+5dydz and V ( d x , d y , d z ) = 2 d x V ( d x , d y , d z ) = 2 d x V((:dx,dy,dz:))=2dx-\mathrm{V}(\langle d x, d y, d z\rangle)=2 d x-V(dx,dy,dz)=2dx d y + d z d y + d z dy+dzd y+d zdy+dz. Find constants c 1 , c 2 c 1 , c 2 c_(1),c_(2)c_{1}, c_{2}c1,c2 and c 3 c 3 c_(3)c_{3}c3, such that
ω ν = c 1 d x d y + c 2 d y d z + c 3 d x d z ω ν = c 1 d x d y + c 2 d y d z + c 3 d x d z omega^^nu=c_(1)dx^^dy+c_(2)dy^^dz+c_(3)dx^^dz\omega \wedge \nu=c_{1} d x \wedge d y+c_{2} d y \wedge d z+c_{3} d x \wedge d zων=c1dxdy+c2dydz+c3dxdz
4.17. Express each of the following as the product of two 1 -forms:
  1. 3 d x d y + d y d x 3 d x d y + d y d x 3dx^^dy+dy^^dx3 d x \wedge d y+d y \wedge d x3dxdy+dydx.
  2. d x d y + d x d z d x d y + d x d z dx^^dy+dx^^dzd x \wedge d y+d x \wedge d zdxdy+dxdz.
  3. 3 d x d y + d y d x + d x d z 3 d x d y + d y d x + d x d z 3dx^^dy+dy^^dx+dx^^dz3 d x \wedge d y+d y \wedge d x+d x \wedge d z3dxdy+dydx+dxdz.
  4. d x d y + 3 d z d y + 4 d x d z d x d y + 3 d z d y + 4 d x d z dx^^dy+3dz^^dy+4dx^^dzd x \wedge d y+3 d z \wedge d y+4 d x \wedge d zdxdy+3dzdy+4dxdz.

4.4 2-forms on Tp 3 Tp 3 Tp^(3)\operatorname{Tp}^{\mathbf{3}}Tp3 (optional)

4.18. Find a 2 -form which is not the product of 1 -forms.
In doing this exercise, you may guess that, in fact, all 2-forms on T p 3 T p 3 T_(p)^(3)T_{p}{ }^{3}Tp3 can be written as a product of 1-forms. Let's see a proof of this fact that relies heavily on the geometric interpretations we have developed.
Recall the correspondence introduced above between vectors and 1 -forms. If a = a 1 d x + a 2 d y + a 3 d z a = a 1 d x + a 2 d y + a 3 d z a=a_(1)dx+a_(2)dy+a_(3)dza=a_{1} d x+a_{2} d y+a_{3} d za=a1dx+a2dy+a3dz then we let a = a 1 , a 2 , a 3 a = a 1 , a 2 , a 3 (:a:)=(:a_(1),a_(2),a_(3):)\langle a\rangle=\left\langle a_{1}, a_{2}, a_{3}\right\ranglea=a1,a2,a3. If V V VVV is a vector, then we let v 1 v 1 (:v:)^(-1)\langle v\rangle^{-1}v1 be the corresponding 1 -form.
We now prove two lemmas:
Lemma 1. If a a aaa and β β beta\betaβ are 1 -forms on T p 3 T p 3 T_(p)^(3)T_{p}{ }^{3}Tp3 and V V VVV is a vector in the plane spanned by a a (:a:)\langle a\ranglea and β β (:beta:)\langle\beta\rangleβ, then there is a vector, W W WWW, in this plane such that a β = v 1 w 1 a β = v 1 w 1 a^^beta=(:v:)^(-1)^^(:w:)^(-1)a \wedge \beta=\langle v\rangle^{-1} \wedge\langle w\rangle^{-1}aβ=v1w1.
Proof. The proof of the above lemma relies heavily on the fact that 2 -forms which are the product of 1 -forms are very flexible. The 2form a β a β a^^betaa \wedge \betaaβ takes pairs of vectors, projects them onto the plane spanned by the vectors a a (:a:)\langle a\ranglea and β β (:beta:)\langle\beta\rangleβ, and computes the area of the resulting parallelogram times the area of the parallelogram spanned by a a (:a:)\langle a\ranglea and β β (:beta:)\langle\beta\rangleβ. Note, that for every non-zero scalar c c ccc, the area of the parallelogram spanned by a a (:a:)\langle a\ranglea and β β (:beta:)\langle\beta\rangleβ is the same as the area of the parallelogram spanned by c a c a c(:a:)\mathrm{c}\langle a\rangleca and 1 / c β 1 / c β 1//c(:beta:)1 / c\langle\beta\rangle1/cβ. (This is the same thing as saying that a β = c a 1 / c β a β = c a 1 / c β a^^beta=ca^^1//c betaa \wedge \beta=c a \wedge 1 / c \betaaβ=ca1/cβ.) The important point here is that we can scale one of the 1 -forms as much as we want at the expense of the other and get the same 2 -form as a product.
Another thing we can do is apply a rotation to the pair of vectors 〈 a) and β β (:beta:)\langle\beta\rangleβ in the plane which they determine. As the area of the
parallelogram spanned by these two vectors is unchanged by rotation, their product still determines the same 2 -form. In particular, suppose V V VVV is any vector in the plane spanned by a a (:a:)\langle a\ranglea and β β (:beta:)\langle\beta\rangleβ. Then we can rotate a a (:a:)\langle a\ranglea and β β (:beta:)\langle\beta\rangleβ to a a (:a:)\langle a\ranglea and β β (:beta:)\langle\beta\rangleβ so that c d = V c d = V c(:d^('):)=Vc\left\langle d^{\prime}\right\rangle=Vcd=V, for some scalar c c ccc. We can then replace the pair ( a , β ) ( a , β ) (:(a:),(:beta:))\langle(a\rangle,\langle\beta\rangle)(a,β) with the pair ( c a c a c(:a:)c\langle a\rangleca, 1 / c β ) = ( V , 1 / c β ) 1 / c β = ( V , 1 / c β ) {:1//c(:beta^('):))=(V,1//c(:beta:))\left.1 / c\left\langle\beta^{\prime}\right\rangle\right)=(V, 1 / c\langle\beta\rangle)1/cβ)=(V,1/cβ). To complete the proof, let W = 1 / c β W = 1 / c β W=1//c(:beta^('):)W=1 / c\left\langle\beta^{\prime}\right\rangleW=1/cβ.
Lemma 2. If ω 1 = a 1 β 1 ω 1 = a 1 β 1 omega_(1)=a_(1)^^beta_(1)\omega_{1}=a_{1} \wedge \beta_{1}ω1=a1β1 and ω 2 = a 2 β 2 ω 2 = a 2 β 2 omega_(2)=a_(2)^^beta_(2)\omega_{2}=a_{2} \wedge \beta_{2}ω2=a2β2 are 2 -forms on T p 3 T p 3 T_(p)^(3)T_{p}{ }^{3}Tp3, then there exist 1-forms, a 3 a 3 a_(3)a_{3}a3 and β 3 β 3 beta_(3)\beta_{3}β3, such that ω 1 + ω 2 = a 3 β 3 ω 1 + ω 2 = a 3 β 3 omega_(1)+omega_(2)=a_(3)^^beta_(3)\omega_{1}+\omega_{2}=a_{3} \wedge \beta_{3}ω1+ω2=a3β3. Proof. Let's examine the sum, a 1 β 1 + a 2 β 2 a 1 β 1 + a 2 β 2 a_(1)^^beta_(1)+a_(2)^^beta_(2)a_{1} \wedge \beta_{1}+a_{2} \wedge \beta_{2}a1β1+a2β2. Our first case is that the plane spanned by the pair ( a 1 , β 1 ) a 1 , β 1 ((:a_(1):),(:beta_(1):))\left(\left\langle a_{1}\right\rangle,\left\langle\beta_{1}\right\rangle\right)(a1,β1) is the same as the plane spanned by the pair, ( a 2 , β 2 ) a 2 , β 2 ((:a_(2):),(:beta_(2):))\left(\left\langle a_{2}\right\rangle,\left\langle\beta_{2}\right\rangle\right)(a2,β2). In this case it must be that a 1 β 1 = C a 2 β 2 a 1 β 1 = C a 2 β 2 a_(1)^^beta_(1)=Ca_(2)^^beta_(2)a_{1} \wedge \beta_{1}=C a_{2} \wedge \beta_{2}a1β1=Ca2β2, and hence, a 1 β 1 + a 2 β 2 = ( 1 + C ) a 1 β 1 a 1 β 1 + a 2 β 2 = ( 1 + C ) a 1 β 1 a_(1)^^beta_(1)+a_(2)^^beta_(2)=(1+C)a_(1)^^beta_(1)a_{1} \wedge \beta_{1}+a_{2} \wedge \beta_{2}=(1+C) a_{1} \wedge \beta_{1}a1β1+a2β2=(1+C)a1β1.
If these two planes are not the same, then they intersect in a line. Let V V VVV be a vector contained in this line. Then by the preceding lemma there are 1-forms γ γ gamma\gammaγ and γ γ gamma\gammaγ such that a 1 β 1 = V 1 γ a 1 β 1 = V 1 γ a_(1)^^beta_(1)=(:V:)^(-1)^^gammaa_{1} \wedge \beta_{1}=\langle V\rangle^{-1} \wedge \gammaa1β1=V1γ and a 2 β 2 = v 1 γ a 2 β 2 = v 1 γ a_(2)^^beta_(2)=(:v:)^(-1)^^gammaa_{2} \wedge \beta_{2}=\langle v\rangle^{-1} \wedge \gammaa2β2=v1γ. Hence,
α 1 β 1 + α 2 β 2 = V 1 γ + V 1 γ = V 1 ( γ + γ ) α 1 β 1 + α 2 β 2 = V 1 γ + V 1 γ = V 1 γ + γ alpha_(1)^^beta_(1)+alpha_(2)^^beta_(2)=(:V:)^(-1)^^gamma+(:V:)^(-1)^^gamma^(')=(:V:)^(-1)^^(gamma+gamma^('))\alpha_{1} \wedge \beta_{1}+\alpha_{2} \wedge \beta_{2}=\langle V\rangle^{-1} \wedge \gamma+\langle V\rangle^{-1} \wedge \gamma^{\prime}=\langle V\rangle^{-1} \wedge\left(\gamma+\gamma^{\prime}\right)α1β1+α2β2=V1γ+V1γ=V1(γ+γ)
Now note that any 2 -form is the sum of products of 1 -forms. Hence, this last lemma implies that any 2 -form on T p 3 T p 3 T_(p)^(3)T_{p}{ }^{3}Tp3 is a product of 1 -forms. In other words:
Every 2-form on T p 3 T p 3 T_(p)^(3)T_{p}{ }^{3}Tp3 projects pairs of vectors onto some plane and returns the area of the resulting parallelogram, scaled by some constant.
This fact is precisely why all of classical vector calculus works. We explore this in the next few exercises, and further in Section 7.3.
4.19. Use the above geometric interpretation of the action of a 2 form on T p 3 T p 3 T_(p)^(3)T_{p}{ }^{3}Tp3 to justify the following statement: For every 2-form ω ω omega\omegaω on T p 3 T p 3 T_(p)^(3)T_{p}{ }^{3}Tp3 there are non-zero vectors V 1 V 1 V_(1)V_{1}V1 and V 2 V 2 V_(2)V_{2}V2 such that V 1 V 1 V_(1)V_{1}V1 is not a multiple of V 2 V 2 V_(2)V_{2}V2, but ω ( V 1 , V 2 ) = 0 ω V 1 , V 2 = 0 omega(V_(1),V_(2))=0\omega\left(V_{1}, V_{2}\right)=0ω(V1,V2)=0.

4.20. Does Problem 4.19 generalize to higher dimensions?

4.21. Show that if ω ω omega\omegaω is a 2-form on T p 3 T p 3 T_(p)^(3)T_{p}{ }^{3}Tp3, then there is a line / in T p T p T_(p)T_{p}Tp 3 3 ^(3){ }^{3}3 such that if the plane spanned by V 1 V 1 V_(1)V_{1}V1 and V 2 V 2 V_(2)V_{2}V2 contains I I III, then ω ω omega\omegaω ( V 1 , V 2 ) = 0 V 1 , V 2 = 0 (V_(1),V_(2))=0\left(V_{1}, V_{2}\right)=0(V1,V2)=0.
Note that the conditions of Problem 4.21 are satisfied when the vectors that are perpendicular to both V 1 V 1 V_(1)V_{1}V1 and V 2 V 2 V_(2)V_{2}V2 are also perpendicular to l l lll.
4.22. Show that if all you know about V 1 V 1 V_(1)V_{1}V1 and V 2 V 2 V_(2)V_{2}V2 is that they are vectors in T p 3 T p 3 T_(p)^(3)T_{p}{ }^{3}Tp3 that span a parallelogram of area A A AAA, then the value of ω ( V 1 , V 2 ) ω V 1 , V 2 omega(V_(1),V_(2))\omega\left(V_{1}, V_{2}\right)ω(V1,V2) is maximized when V 1 V 1 V_(1)V_{1}V1 and V 2 V 2 V_(2)V_{2}V2 are perpendicular to the line / of Problem 4.21.
Note that the conditions of this exercise are satisfied when the vectors perpendicular to V 1 V 1 V_(1)V_{1}V1 and V 2 V 2 V_(2)V_{2}V2 are parallel to I I III.
4.23. Let N N NNN be a vector perpendicular to V 1 V 1 V_(1)V_{1}V1 and V 2 V 2 V_(2)V_{2}V2 in T p 3 T p 3 T_(p)^(3)T_{p}{ }^{3}Tp3 whose length is precisely the area of the parallelogram spanned by these two vectors. Show that there is a vector V ω V ω V_(omega)V_{\omega}Vω in the line / of Problem 4.21 such that the value of ω ( V 1 , V 2 ) ω V 1 , V 2 omega(V_(1),V_(2))\omega\left(V_{1}, V_{2}\right)ω(V1,V2) is precisely V ω N V ω N V_(omega)*NV_{\omega} \cdot NVωN.
Remark. You may have learned that the vector N N NNN of the previous exercise is precisely the cross product of V 1 V 1 V_(1)V_{1}V1 and V 2 V 2 V_(2)V_{2}V2. Hence, the previous problem implies that if ω ω omega\omegaω is a 2-form on T p 3 T p 3 T_(p)^(3)T_{p}{ }^{3}Tp3 then there is a vector V ω V ω V_(omega)V_{\omega}Vω such that ω ( V 1 , V 2 ) = V ω ( V 1 × V 2 ) ω V 1 , V 2 = V ω V 1 × V 2 omega(V_(1),V_(2))=V_(omega)*(V_(1)xxV_(2))\omega\left(V_{1}, V_{2}\right)=V_{\omega} \cdot\left(V_{1} \times V_{2}\right)ω(V1,V2)=Vω(V1×V2).
4.24. Show that if ω = F x d y d z F y d x d z + F z d x d y ω = F x d y d z F y d x d z + F z d x d y omega=F_(x)dy^^dz-F_(y)dx^^dz+F_(z)dx^^dy\omega=F_{x} d y \wedge d z-F_{y} d x \wedge d z+F_{z} d x \wedge d yω=FxdydzFydxdz+Fzdxdy, then V V VVV ω = F x F y F z ω = F x F y F z omega=(:F_(x^('))F_(y^('))F_(z^(')):)\omega=\left\langle F_{x^{\prime}} F_{y^{\prime}} F_{z^{\prime}}\right\rangleω=FxFyFz.

4.5 2-forms and 3-forms on Tp 4 4 ^(4){ }^{4}4 (optional)

Many of the techniques of the previous section can be used to prove results about 2- and 3-forms on T p 4 T p 4 T_(p)^(4)T_{p}{ }^{4}Tp4.
4.25. Show that any 3 -form on T p 4 T p 4 T_(p)^(4)T_{p}{ }^{4}Tp4 can be written as the product of three 1-forms. (Hint. Two three-dimensional subspaces of T p 4 T p 4 T_(p)^(4)T_{p}{ }^{4}Tp4 must meet in at least a line.)
We now give away an answer to Problem 4.18. Let ω = d x d y + ω = d x d y + omega=dx^^dy+\omega=d x \wedge d y+ω=dxdy+ d z d w d z d w dz^^dwd z \wedge d wdzdw. Then an easy computation shows that ω ω = 2 d x d y ω ω = 2 d x d y omega^^omega=2dx^^dy\omega \wedge \omega=2 d x \wedge d yωω=2dxdy d z d w d z d w ^^dz^^dw\wedge d z \wedge d wdzdw. But if ω ω omega\omegaω were equal to a β a β a^^betaa \wedge \betaaβ, for some 1 -forms a a aaa and β β beta\betaβ, then ω ω ω ω omega^^omega\omega \wedge \omegaωω would be zero (why?). This argument shows that in general, if ω ω omega\omegaω is any 2 -form such that ω ω 0 ω ω 0 omega^^omega!=0\omega \wedge \omega \neq 0ωω0, then ω ω omega\omegaω cannot be written as the product of 1 -forms.
4.26. Let ω ω omega\omegaω be a 2 -form on T p T p T_(p)T_{p}Tp. Show that ω ω omega\omegaω can be written as the sum of exactly two products. That is, ω = a β + δ γ ω = a β + δ γ omega=a^^beta+delta^^gamma\omega=a \wedge \beta+\delta \wedge \gammaω=aβ+δγ. (Hint. Given three planes in T p 4 T p 4 T_(p)^(4)T_{p}{ }^{4}Tp4 there are at least two of them that intersect in more than a point.)
Above, we saw that if ω ω omega\omegaω is a 2-form such that ω ω 0 ω ω 0 omega^^omega!=0\omega \wedge \omega \neq 0ωω0, then ω ω omega\omegaω is not the product of 1 -forms. We now use the previous exercise to show the converse:
Theorem 1. If ω ω omega\omegaω is a 2 -form on T p T p T_(p)T_{p}Tp such that ω ω = 0 ω ω = 0 omega^^omega=0\omega \wedge \omega=0ωω=0, then ω ω omega\omegaω can be written as the product of two 1 -forms.
Our proof of this again relies heavily on the geometry of the situation. By the previous exercise, ω = a β + δ γ ω = a β + δ γ omega=a^^beta+delta^^gamma\omega=a \wedge \beta+\delta \wedge \gammaω=aβ+δγ. A short computation then shows
ω ω = 2 α β δ γ . ω ω = 2 α β δ γ . omega^^omega=2alpha^^beta^^delta^^gamma.\omega \wedge \omega=2 \alpha \wedge \beta \wedge \delta \wedge \gamma .ωω=2αβδγ.
If this 4-form is the zero 4-form, then it must be the case that the (4-dimensional) volume of the parallelepiped spanned by a , β , δ a , β , δ (:a:),(:beta:),(:delta:)\langle a\rangle,\langle\beta\rangle,\langle\delta\ranglea,β,δ and y y (:y:)\langle y\rangley is zero. This, in turn, implies that the plane spanned by a a (:a:)\langle a\ranglea and β β (:beta:)\langle\beta\rangleβ meets the plane spanned by δ δ (:delta:)\langle\delta\rangleδ and γ γ (:gamma:)\langle\gamma\rangleγ in at least a line (show this!). Call such an intersection line L L LLL.
As in the previous section, we can now rotate a a (:a:)\langle a\ranglea and β β (:beta:)\langle\beta\rangleβ, in the plane they span, to vectors a a (:a^('):)\left\langle a^{\prime}\right\ranglea and β β (:beta^('):)\left\langle\beta^{\prime}\right\rangleβ such that a a (:a^('):)\left\langle a^{\prime}\right\ranglea lies in the line L L LLL. The 2 -form a β a β a^(')^^beta^(')a^{\prime} \wedge \beta^{\prime}aβ must equal a β a β a^^betaa \wedge \betaaβ since they determine the same plane, and span a parallelogram of the same area. Similarly, we rotate δ δ (:delta:)\langle\delta\rangleδ and γ γ (:gamma:)\langle\gamma\rangleγ to vectors δ δ (:delta^('):)\left\langle\delta^{\prime}\right\rangleδ and γ γ (:gamma:)\langle\gamma\rangleγ such that δ L δ L (:delta^('):)sub L\left\langle\delta^{\prime}\right\rangle \subset LδL. It follows that δ γ = δ γ δ γ = δ γ delta^^gamma=delta^(')^^gamma\delta \wedge \gamma=\delta^{\prime} \wedge \gammaδγ=δγ.
Since a a (:a^('):)\left\langle a^{\prime}\right\ranglea and δ δ (:delta^('):)\left\langle\delta^{\prime}\right\rangleδ lie on the same line, there is a constant c c ccc such that c a = δ c a = δ ca^(')=delta^(')c a^{\prime}=\delta^{\prime}ca=δ. We now put all of this information together:
ω = α β + δ γ = α β + δ γ = ( c α ) ( 1 c β ) + δ γ = δ ( 1 c β ) + δ γ = δ ( 1 c β + γ ) . ω = α β + δ γ = α β + δ γ = c α 1 c β + δ γ = δ 1 c β + δ γ = δ 1 c β + γ . {:[omega=alpha^^beta+delta^^gamma],[=alpha^(')^^beta^(')+delta^(')^^gamma^(')],[=(calpha^('))^^((1)/(c)beta^('))+delta^(')^^gamma^(')],[=delta^(')^^((1)/(c)beta^('))+delta^(')^^gamma^(')],[=delta^(')^^((1)/(c)beta^(')+gamma^(')).]:}\begin{aligned} \omega & =\alpha \wedge \beta+\delta \wedge \gamma \\ & =\alpha^{\prime} \wedge \beta^{\prime}+\delta^{\prime} \wedge \gamma^{\prime} \\ & =\left(c \alpha^{\prime}\right) \wedge\left(\frac{1}{c} \beta^{\prime}\right)+\delta^{\prime} \wedge \gamma^{\prime} \\ & =\delta^{\prime} \wedge\left(\frac{1}{c} \beta^{\prime}\right)+\delta^{\prime} \wedge \gamma^{\prime} \\ & =\delta^{\prime} \wedge\left(\frac{1}{c} \beta^{\prime}+\gamma^{\prime}\right) . \end{aligned}ω=αβ+δγ=αβ+δγ=(cα)(1cβ)+δγ=δ(1cβ)+δγ=δ(1cβ+γ).

4.6 n-forms

Let's think a little more about our multiplication operator, Λ Λ Lambda\LambdaΛ. If it is really going to be anything like multiplication, we should be able to take three 1-forms, ω , vv ω , vv omega,vv\omega, \mathrm{vv}ω,vv and Ψ Ψ Psi\PsiΨ, and form the product ω Ψ ω Ψ omega^^vv^^Psi\omega \wedge \vee \wedge \PsiωΨ. How can we define this? A first guess might be to say that ω ω omega^^vv^^\omega \wedge \vee \wedgeω Ψ = ω ( Ψ ) Ψ = ω ( Ψ ) Psi=omega^^(vv^^Psi)\Psi=\omega \wedge(\vee \wedge \Psi)Ψ=ω(Ψ), but Ψ Ψ vv^^Psi\vee \wedge \PsiΨ is a 2-form and we have not defined the product of a 2-form and a 1-form. We take a different approach and define ω Ψ ω Ψ omega^^vv^^Psi\omega \wedge \vee \wedge \PsiωΨ directly.
This is completely analogous to the previous section. ω , ω , omega,vv\omega, \veeω, and Ψ Ψ Psi\PsiΨ each act on a vector, V V VVV, to give three numbers. In other words, they can be thought of as coordinate functions. We say the coordinates of V V VVV are [ ω ( V ) , v ( V ) , Ψ ( V ) ] [ ω ( V ) , v ( V ) , Ψ ( V ) ] [omega(V),v(V),Psi(V)][\omega(V), v(V), \Psi(V)][ω(V),v(V),Ψ(V)]. Hence, if we have three vectors, V 1 , V 2 V 1 , V 2 V_(1),V_(2)V_{1}, V_{2}V1,V2 and V 3 V 3 V_(3)V_{3}V3, we can compute the [ ω , V , Ψ ] [ ω , V , Ψ ] [omega,V,Psi][\omega, V, \Psi][ω,V,Ψ] coordinates of each. This gives us three new vectors. The signed volume of the parallelepiped which they span is what we define to be the value of ω Ψ ( V 1 ω Ψ V 1 omega^^vv^^Psi(V_(1):}\omega \wedge \vee \wedge \Psi\left(V_{1}\right.ωΨ(V1 , V 2 , V 3 , V 2 , V 3 ,V_(2),V_(3), V_{2}, V_{3},V2,V3 ).
There is no reason to stop at three dimensions. Suppose ω 1 , ω 2 ω 1 , ω 2 omega_(1),omega_(2)\omega_{1}, \omega_{2}ω1,ω2, , ω n , ω n dots,omega_(n)\ldots, \omega_{n},ωn are 1 -forms and V 1 , V 2 , , V n V 1 , V 2 , , V n V_(1),V_(2),dots,V_(n)V_{1}, V_{2}, \ldots, V_{n}V1,V2,,Vn are vectors. Then we define the value of to be the signed ( n n nnn-dimensional) volume of the parallelepiped spanned by the vectors [ ω 1 ( V i ) , ω 2 ( V i ) , , ω n ( V i ) ] ω 1 V i , ω 2 V i , , ω n V i [omega_(1)(V_(i)),omega_(2)(V_(i)),dots,omega_(n)(V_(i))]\left[\omega_{1}\left(V_{i}\right), \omega_{2}\left(V_{i}\right), \ldots, \omega_{n}\left(V_{i}\right)\right][ω1(Vi),ω2(Vi),,ωn(Vi)]. Algebraically,
ω 1 ω 2 ω n ( V 1 , V 2 , , V n ) = | ω 1 ( V 1 ) ω 1 ( V 2 ) ω 1 ( V n ) ω 2 ( V 1 ) ω 2 ( V 2 ) ω 2 ( V n ) ω n ( V 1 ) ω n ( V 2 ) ω n ( V n ) | . ω 1 ω 2 ω n V 1 , V 2 , , V n = ω 1 V 1 ω 1 V 2 ω 1 V n ω 2 V 1 ω 2 V 2 ω 2 V n ω n V 1 ω n V 2 ω n V n . omega_(1)^^omega_(2)^^dots^^omega_(n)(V_(1),V_(2),dots,V_(n))=|[omega_(1)(V_(1)),omega_(1)(V_(2)),dots,omega_(1)(V_(n))],[omega_(2)(V_(1)),omega_(2)(V_(2)),dots,omega_(2)(V_(n))],[vdots,vdots,,vdots],[omega_(n)(V_(1)),omega_(n)(V_(2)),dots,omega_(n)(V_(n))]|.\omega_{1} \wedge \omega_{2} \wedge \ldots \wedge \omega_{n}\left(V_{1}, V_{2}, \ldots, V_{n}\right)=\left|\begin{array}{cccc} \omega_{1}\left(V_{1}\right) & \omega_{1}\left(V_{2}\right) & \ldots & \omega_{1}\left(V_{n}\right) \\ \omega_{2}\left(V_{1}\right) & \omega_{2}\left(V_{2}\right) & \ldots & \omega_{2}\left(V_{n}\right) \\ \vdots & \vdots & & \vdots \\ \omega_{n}\left(V_{1}\right) & \omega_{n}\left(V_{2}\right) & \ldots & \omega_{n}\left(V_{n}\right) \end{array}\right| .ω1ω2ωn(V1,V2,,Vn)=|ω1(V1)ω1(V2)ω1(Vn)ω2(V1)ω2(V2)ω2(Vn)ωn(V1)ωn(V2)ωn(Vn)|.
Note that, just as in Problem 4.12, if a , β a , β a,betaa, \betaa,β and γ γ gamma\gammaγ are 1-forms on T p T p T_(p)T_{p}Tp 3, then a β γ ( V 1 , V 2 , V 3 ) a β γ V 1 , V 2 , V 3 a^^beta^^gamma(V_(1),V_(2),V_(3))a \wedge \beta \wedge \gamma\left(V_{1}, V_{2}, V_{3}\right)aβγ(V1,V2,V3) is the (signed) volume of the parallelepiped spanned by V 1 , V 2 V 1 , V 2 V_(1),V_(2)V_{1}, V_{2}V1,V2 and V 3 V 3 V_(3)V_{3}V3 times the volume of the parallelepiped spanned by a , β a , β (:a:),(:beta:)\langle a\rangle,\langle\beta\ranglea,β and γ γ (:gamma:)\langle\gamma\rangleγ.
4.27. Suppose ω ω omega\omegaω is a 2-form on T p 3 T p 3 T_(p)^(3)T_{p}{ }^{3}Tp3 and v v vvv is a 1 -form on T p 3 T p 3 T_(p)^(3)T_{p}{ }^{3}Tp3. Show that if ω v = 0 ω v = 0 omega^^v=0\omega \wedge v=0ωv=0, then there is a 1-form γ γ gamma\gammaγ such that ω = ω = omega=vv^^\omega=\vee \wedgeω= γ γ gamma\gammaγ. (Hint. Combine the above geometric interpretation of a 3-form, which is the product of 1 -forms on T p R 3 T p R 3 T_(p)R^(3)T_{p} R^{3}TpR3, with the results of Section 4.4.)
It follows from linear algebra that if we swap any two rows or columns of this matrix, the sign of the result flips. Hence, if the n n nnn tuple, V = ( V i 1 , V i 2 , , V i n ) V = V i 1 , V i 2 , , V i n V^(')=(V_(i1),V_(i2),dots,V_(in))\mathbf{V}^{\prime}=\left(V_{i 1}, V_{i 2}, \ldots, V_{i n}\right)V=(Vi1,Vi2,,Vin) is obtained from V = ( V 1 , V 2 , , V n ) V = V 1 , V 2 , , V n V=(V_(1),V_(2),dots,V_(n))\mathbf{V}=\left(V_{1}, V_{2}, \ldots, V_{n}\right)V=(V1,V2,,Vn) by an even number of exchanges, then the sign of ω 1 ω 2 ω 1 ω 2 omega_(1)^^omega_(2)^^dots^^\omega_{1} \wedge \omega_{2} \wedge \ldots \wedgeω1ω2 ω n ( V ) ω n V omega_(n)(V^('))\omega_{n}\left(\mathbf{V}^{\prime}\right)ωn(V) will be the same as the sign of ω 1 ω 2 ω n ( V ) ω 1 ω 2 ω n ( V ) omega_(1)^^omega_(2)^^dots^^omega_(n)(V)\omega_{1} \wedge \omega_{2} \wedge \ldots \wedge \omega_{n}(\mathbf{V})ω1ω2ωn(V). If the number of exchanges is odd, then the sign is opposite. We sum this up by saying that the n n nnn-form, ω 1 ω 2 ω n ω 1 ω 2 ω n omega_(1)^^omega_(2)^^dots^^omega_(n)\omega_{1} \wedge \omega_{2} \wedge \ldots \wedge \omega_{n}ω1ω2ωn is alternating.
The wedge product of 1 -forms is also multilinear, in the following sense:
ω 1 ω 2 ω n ( V 1 , , V i + V i , , V n ) = ω 1 ω 2 ω n ( V 1 , , V i , , V n ) + ω 1 ω 2 ω n V 1 , , V i + V i , , V n = ω 1 ω 2 ω n V 1 , , V i , , V n + omega_(1)^^omega_(2)^^dots^^omega_(n)(V_(1),dots,V_(i)+V_(i)^('),dots,V_(n))=omega_(1)^^omega_(2)^^dots^^omega_(n)(V_(1),dots,V_(i),dots,V_(n))+\omega_{1} \wedge \omega_{2} \wedge \ldots \wedge \omega_{n}\left(V_{1}, \ldots, V_{i}+V_{i}^{\prime}, \ldots, V_{n}\right)=\omega_{1} \wedge \omega_{2} \wedge \ldots \wedge \omega_{n}\left(V_{1}, \ldots, V_{i}, \ldots, V_{n}\right)+ω1ω2ωn(V1,,Vi+Vi,,Vn)=ω1ω2ωn(V1,,Vi,,Vn)+ ω 1 ω 2 ω n ( V 1 , , V i , , V n ) ω 1 ω 2 ω n V 1 , , V i , , V n omega_(1)^^omega_(2)^^dots^^omega_(n)(V_(1),dots,V_(i)^('),dots,V_(n))\omega_{1} \wedge \omega_{2} \wedge \ldots \wedge \omega_{n}\left(V_{1}, \ldots, V_{i}^{\prime}, \ldots, V_{n}\right)ω1ω2ωn(V1,,Vi,,Vn),
and
ω 1 ω 2 ω n ( V 1 , , c V i , , V n ) = c ω 1 ω 2 ω n ( V 1 ω 1 ω 2 ω n V 1 , , c V i , , V n = c ω 1 ω 2 ω n V 1 omega_(1)^^omega_(2)^^dots^^omega_(n)(V_(1),dots,cV_(i),dots,V_(n))=comega_(1)^^omega_(2)^^dots^^omega_(n)(V_(1):}\omega_{1} \wedge \omega_{2} \wedge \ldots \wedge \omega_{n}\left(V_{1}, \ldots, c V_{i}, \ldots, V_{n}\right)=c \omega_{1} \wedge \omega_{2} \wedge \ldots \wedge \omega_{n}\left(V_{1}\right.ω1ω2ωn(V1,,cVi,,Vn)=cω1ω2ωn(V1, , V i , , V n , V i , , V n dots,V_(i),dots,V_(n)\ldots, V_{i}, \ldots, V_{n},Vi,,Vn ), for all i i iii and any real number, c c ccc.
In general, we define an n n nnn-form to be any alternating, multilinear real-valued function which acts on n n nnn-tuples of vectors.
4.28. Prove the following geometric interpretation: (Hint. All of the steps are completely analogous to those in the last section.)
An m-form on T p n T p n T_(p)^(n)T_{p}{ }^{n}Tpn can be thought of as a function which takes the parallelepiped spanned by m m mmm vectors, projects it onto each of the m m mmm dimensional coordinate planes, computes the resulting areas, multiplies each by some constant, and adds the results.
4.29. How many numbers do you need to give to specify a 5 -form on T p 10 T p 10 T_(p)^(10)T_{p}{ }^{10}Tp10 ?
We turn now to the simple case of an n n nnn-form on T p n n T p n n T_(p)n^(n)T_{p} n^{n}Tpnn. Notice that there is only one n n nnn-dimensional coordinate plane in this space, namely, the space itself. Such a form, evaluated on an n n nnn-tuple of vectors, must therefore give the n n nnn-dimensional volume of the parallelepiped which it spans, multiplied by some constant. For this reason such a form is called a volume form (in 2-dimensions, an area form).
Example 15. Consider the forms, ω = d x + 2 d y d z , v = 3 d x d y + ω = d x + 2 d y d z , v = 3 d x d y + omega=dx+2dy-dz,v=3dx-dy+\omega=d x+2 d y-d z, \mathrm{v}=3 d x-d y+ω=dx+2dydz,v=3dxdy+ d z d z dzd zdz and Ψ = d x 3 d y + d z Ψ = d x 3 d y + d z Psi=-dx-3dy+dz\Psi=-d x-3 d y+d zΨ=dx3dy+dz, on T p 3 T p 3 T_(p)^(3)T_{p}{ }^{3}Tp3. By the above argument ω v ω v omega^^v\omega \wedge vωv Ψ Ψ ^^Psi\wedge \PsiΨ must be a volume form. But which volume form is it? One way to tell is to compute its value on a set of vectors which we know span a parallelepiped of volume one, namely 1 , 0 , 0 , 0 , 1 , 0 1 , 0 , 0 , 0 , 1 , 0 (:1,0,0:),(:0,1,0:)\langle 1,0,0\rangle,\langle 0,1,0\rangle1,0,0,0,1,0 and (:\langle 0 , 0 , 1 0 , 0 , 1 0,0,1:)0,0,1\rangle0,0,1. This will tell us how much the form scales volume.
ω v ψ ( 1 , 0 , 0 , 0 , 1 , 0 , 0 , 0 , 1 ) = | 1 2 1 3 1 1 1 3 1 | = 4 ω v ψ ( 1 , 0 , 0 , 0 , 1 , 0 , 0 , 0 , 1 ) = 1      2      1 3      1      1 1      3      1 = 4 omega^^v^^psi((:1,0,0:),(:0,1,0:),(:0,0,1:))=|[1,2,-1],[3,-1,1],[-1,-3,1]|=4\omega \wedge v \wedge \psi(\langle 1,0,0\rangle,\langle 0,1,0\rangle,\langle 0,0,1\rangle)=\left|\begin{array}{rrr} 1 & 2 & -1 \\ 3 & -1 & 1 \\ -1 & -3 & 1 \end{array}\right|=4ωvψ(1,0,0,0,1,0,0,0,1)=|121311131|=4
So, ω Ψ ω Ψ omega^^vv^^Psi\omega \wedge \vee \wedge \PsiωΨ must be the same as the form 4 d x d y d z 4 d x d y d z 4dx^^dy^^dz4 d x \wedge d y \wedge d z4dxdydz.
4.30. Let ω ( d x , d y , d z ) = d x + 5 d y d z , v ( d x , d y , d z ) = 2 d x d y ω ( d x , d y , d z ) = d x + 5 d y d z , v ( d x , d y , d z ) = 2 d x d y omega((:dx,dy,dz:))=dx+5dy-dz,v((:dx,dy,dz:))=2dx-dy\omega(\langle d x, d y, d z\rangle)=d x+5 d y-d z, \mathrm{v}(\langle d x, d y, d z\rangle)=2 d x-d yω(dx,dy,dz)=dx+5dydz,v(dx,dy,dz)=2dxdy + d z + d z +dz+d z+dz and y ( d x , d y , d z ) = d x + d y + 2 d z y ( d x , d y , d z ) = d x + d y + 2 d z y((:dx,dy,dz)=-dx+dy+2dzy(\langle d x, d y, d z)=-d x+d y+2 d zy(dx,dy,dz)=dx+dy+2dz.
  1. If V 1 = 1 , 0 , 2 , V 2 = 1 , 1 , 2 V 1 = 1 , 0 , 2 , V 2 = 1 , 1 , 2 V_(1)=(:1,0,2:),V_(2)=(:1,1,2:)V_{1}=\langle 1,0,2\rangle, V_{2}=\langle 1,1,2\rangleV1=1,0,2,V2=1,1,2 and V 3 = 0 , 2 , 3 V 3 = 0 , 2 , 3 V_(3)=(:0,2,3:)V_{3}=\langle 0,2,3\rangleV3=0,2,3, compute ω ω omega\omegaω γ ( V 1 , V 2 , V 3 ) γ V 1 , V 2 , V 3 ^^vv^^gamma(V_(1),V_(2),V_(3))\wedge \vee \wedge \gamma\left(V_{1}, V_{2}, V_{3}\right)γ(V1,V2,V3).
  2. Find a constant, c c ccc, such that ω γ = c d x d y d z ω γ = c d x d y d z omega^^vv^^gamma=cdx^^dy^^dz\omega \wedge \vee \wedge \gamma=c d x \wedge d y \wedge d zωγ=cdxdydz.
  3. Let a = 3 d x d y + 2 d y d z d x d z a = 3 d x d y + 2 d y d z d x d z a=3dx^^dy+2dy^^dz-dx^^dza=3 d x \wedge d y+2 d y \wedge d z-d x \wedge d za=3dxdy+2dydzdxdz. Find a constant, c c ccc, such that a γ = c d x d y d z a γ = c d x d y d z a^^gamma=cdx^^dy^^dza \wedge \gamma=c d x \wedge d y \wedge d zaγ=cdxdydz.
    4.31. Simplify:
d x d y d z + d x d z d y + d y d z d x + d y d x d y d x d y d z + d x d z d y + d y d z d x + d y d x d y dx^^dy^^dz+dx^^dz^^dy+dy^^dz^^dx+dy^^dx^^dyd x \wedge d y \wedge d z+d x \wedge d z \wedge d y+d y \wedge d z \wedge d x+d y \wedge d x \wedge d ydxdydz+dxdzdy+dydzdx+dydxdy
4.32. Let ω ω omega\omegaω be an n n nnn-form and vv\vee an m m mmm-form.
  1. Show that
ω ν = ( 1 ) n m ν ω . ω ν = ( 1 ) n m ν ω . omega^^nu=(-1)^(nm)nu^^omega.\omega \wedge \nu=(-1)^{n m} \nu \wedge \omega .ων=(1)nmνω.
  1. Use this to show that if n n nnn is odd then ω ω = 0 ω ω = 0 omega^^omega=0\omega \wedge \omega=0ωω=0.

4.7 Algebraic computation of products

In this section, we break with the spirit of the text briefly. At this point, we have amassed enough algebraic identities that multiplying forms becomes similar to multiplying polynomials. We quickly summarize these identities and work a few examples.
Let ω ω omega\omegaω be an n n nnn-form and v be an m m mmm-form. Then we have the following identities
ω v = ( 1 ) n m v ω ω ω = 0 if n is odd ω ( v + ψ ) = ω v + ω ψ ( v + ψ ) ω = v ω + ψ ω ω v = ( 1 ) n m v ω ω ω = 0  if  n  is odd  ω ( v + ψ ) = ω v + ω ψ ( v + ψ ) ω = v ω + ψ ω {:[omega^^v=(-1)^(nm)v^^omega],[omega^^omega=0" if "n" is odd "],[omega^^(v+psi)=omega^^v+omega^^psi],[(v+psi)^^omega=v^^omega+psi^^omega]:}\begin{aligned} \omega \wedge v & =(-1)^{n m} v \wedge \omega \\ \omega \wedge \omega & =0 \text { if } n \text { is odd } \\ \omega \wedge(v+\psi) & =\omega \wedge v+\omega \wedge \psi \\ (v+\psi) \wedge \omega & =v \wedge \omega+\psi \wedge \omega \end{aligned}ωv=(1)nmvωωω=0 if n is odd ω(v+ψ)=ωv+ωψ(v+ψ)ω=vω+ψω

Example 16.

( x d x + y d y ) ( y d x + x d y ) = x y d x d x + x 2 d x d y + y 2 d y d x + y x d y d y = x 2 d x d y + y 2 d y d x = x 2 d x d y y 2 d x d y = ( x 2 y 2 ) d x d y ( x d x + y d y ) ( y d x + x d y ) = x y d x d x + x 2 d x d y + y 2 d y d x + y x d y d y = x 2 d x d y + y 2 d y d x = x 2 d x d y y 2 d x d y = x 2 y 2 d x d y {:[(xdx+ydy)^^(ydx+xdy)=xydx^^dx+x^(2)dx^^dy+y^(2)dy^^dx],[+yxdy^^dy],[=x^(2)dx^^dy+y^(2)dy^^dx],[=x^(2)dx^^dy-y^(2)dx^^dy],[=(x^(2)-y^(2))dx^^dy]:}\begin{aligned} (x d x+y d y) \wedge(y d x+x d y)= & x y d x \wedge d x+x^{2} d x \wedge d y+y^{2} d y \wedge d x \\ & +y x d y \wedge d y \\ = & x^{2} d x \wedge d y+y^{2} d y \wedge d x \\ = & x^{2} d x \wedge d y-y^{2} d x \wedge d y \\ = & \left(x^{2}-y^{2}\right) d x \wedge d y \end{aligned}(xdx+ydy)(ydx+xdy)=xydxdx+x2dxdy+y2dydx+yxdydy=x2dxdy+y2dydx=x2dxdyy2dxdy=(x2y2)dxdy

Example 17.

( x d x + y d y ) ( x z d x d z + y z d y d z ) = x 2 z d x d x d z + x y z d x d y d z + y x z d y d x d z + y 2 z d y d y d z = x y z d x d y d z + y x z d y d x d z = x y z d x d y d z x y z d x d y d z = 0 . ( x d x + y d y ) ( x z d x d z + y z d y d z ) = x 2 z d x d x d z + x y z d x d y d z + y x z d y d x d z + y 2 z d y d y d z = x y z d x d y d z + y x z d y d x d z = x y z d x d y d z x y z d x d y d z = 0 . {:[(xdx+ydy)^^(xzdx^^dz+yzdy^^dz)],[=x^(2)zdx^^dx^^dz+xyzdx^^dy^^dz],[+yxzdy^^dx^^dz+y^(2)zdy^^dy^^dz],[=xyzdx^^dy^^dz+yxzdy^^dx^^dz],[=xyzdx^^dy^^dz-xyzdx^^dy^^dz],[=0.]:}\begin{aligned} (x d x+y d y) \wedge & (x z d x \wedge d z+y z d y \wedge d z) \\ = & x^{2} z d x \wedge d x \wedge d z+x y z d x \wedge d y \wedge d z \\ & +y x z d y \wedge d x \wedge d z+y^{2} z d y \wedge d y \wedge d z \\ = & x y z d x \wedge d y \wedge d z+y x z d y \wedge d x \wedge d z \\ = & x y z d x \wedge d y \wedge d z-x y z d x \wedge d y \wedge d z \\ = & 0 . \end{aligned}(xdx+ydy)(xzdxdz+yzdydz)=x2zdxdxdz+xyzdxdydz+yxzdydxdz+y2zdydydz=xyzdxdydz+yxzdydxdz=xyzdxdydzxyzdxdydz=0.

4.33. Expand and simplify:

  1. [ ( x y ) d x + ( x + y ) d y + z d z ] [ ( x y ) d x + ( x + y ) d y ] [ ( x y ) d x + ( x + y ) d y + z d z ] [ ( x y ) d x + ( x + y ) d y ] [(x-y)dx+(x+y)dy+zdz]^^[(x-y)dx+(x+y)dy][(x-y) d x+(x+y) d y+z d z] \wedge[(x-y) d x+(x+y) d y][(xy)dx+(x+y)dy+zdz][(xy)dx+(x+y)dy].
  2. ( 2 d x + 3 d y ) ( d x d z ) ( d x + d y + d z ) ( 2 d x + 3 d y ) ( d x d z ) ( d x + d y + d z ) (2dx+3dy)^^(dx-dz)^^(dx+dy+dz)(2 d x+3 d y) \wedge(d x-d z) \wedge(d x+d y+d z)(2dx+3dy)(dxdz)(dx+dy+dz).

5

Differential Forms

5.1 Families of forms

Let us now go back to the example in Chapter 3. In the last section of that chapter, we showed that the integral of a function, f : 3 f : 3 f:^(3)rarrf:{ }^{3} \rightarrowf:3 R R RRR, over a surface parameterized by φ : R R 2 φ : R R 2 varphi:R subR^(2)\varphi: R \subset \mathbb{R}^{2}φ:RR2 is
R f ( ϕ ( r , θ ) ) Area [ ϕ r ( r , θ ) , ϕ θ ( r , θ ) ] d r d θ . R f ( ϕ ( r , θ ) )  Area  ϕ r ( r , θ ) , ϕ θ ( r , θ ) d r d θ . int_(R)f(phi(r,theta))" Area "[(del phi)/(del r)(r,theta),(del phi)/(del theta)(r,theta)]drd theta.\int_{R} f(\phi(r, \theta)) \text { Area }\left[\frac{\partial \phi}{\partial r}(r, \theta), \frac{\partial \phi}{\partial \theta}(r, \theta)\right] d r d \theta .Rf(ϕ(r,θ)) Area [ϕr(r,θ),ϕθ(r,θ)]drdθ.
This gave one motivation for studying differential forms. We wanted to generalize this integral by considering functions other than "Area ( , ) ( , ) (*,**)(\cdot, \cdot \cdot)(,) " that eat pairs of vectors and return numbers. But in this integral the point at which such a pair of vectors is based, changes. In other words, Area ( , ) ( , ) (*,*)(\cdot, \cdot)(,) does not act on T p 3 × T p 3 T p 3 × T p 3 T_(p)^(3)xxT_(p)^(3)T_{p}{ }^{3} \times T_{p}{ }^{3}Tp3×Tp3 for a fixed p p ppp. We can make this point a little clearer by re-examining the above integrand. Note that it is of the form f ( ) f ( ) f(**)f(*)f() Area( ( , ) ( , ) (*,*)(\cdot, \cdot)(,). For a fixed point, ***, of 3 3 ^(3){ }^{3}3, this is an operator on T 3 × T 3 T 3 × T 3 T_(**)^(3)xxT_(**)^(3)T_{*}{ }^{3} \times T_{*}{ }^{3}T3×T3, much like a 2 -form is.
But so far all we have done is to define 2 -forms at fixed points of
3 3 ^(3){ }^{3}3. To really generalize the above integral we must start to consider entire families of 2 -forms, ω p : T p × T p 3 ω p : T p × T p 3 omega_(p):T_(p)xxT_(p)^(3)rarr\omega_{p}: T_{p} \times T_{p}{ }^{3} \rightarrowωp:Tp×Tp3, where p p ppp ranges over all of 3 3 ^(3){ }^{3}3. Of course, for this to be useful such a family must have some "niceness" properties. For one thing, it should be continuous. That is, if p p ppp and q q qqq are close, then ω p ω p omega_(p)\omega_{p}ωp and ω q ω q omega_(q)\omega_{q}ωq should be similar.
An even stronger property is that the family, ω p , ω p , omega_(p,)\omega_{p,}ωp, is differentiable. To see what this means recall that for a fixed p p ppp, a 2 -form ω p ω p omega_(p)\omega_{p}ωp can always be written as a p d x d y + b p d y d z + c p d x d z a p d x d y + b p d y d z + c p d x d z a_(p)dx^^dy+b_(p)dy^^dz+c_(p)dx^^dza_{p} d x \wedge d y+b_{p} d y \wedge d z+c_{p} d x \wedge d zapdxdy+bpdydz+cpdxdz, where a p a p a_(p)a_{p}ap, b p b p b_(p)b_{p}bp, and c p c p c_(p)c_{p}cp are constants. But if we let our choice of p p ppp vary over all of 3 3 ^(3){ }^{3}3, then so will these constants. In other words, a p , b p a p , b p a_(p),b_(p)a_{p}, b_{p}ap,bp and c p c p c_(p)c_{p}cp are all functions from R 3 R 3 R^(3)\mathbb{R}^{3}R3 to . To say that the family, ω p ω p omega_(p)\omega_{p}ωp is differentiable we mean that each of these functions is differentiable. If ω p ω p omega_(p)\omega_{p}ωp is differentiable, then we will refer to it as a differential form. When there can be no confusion we will suppress the subscript, p p ppp. Example 18. ω = x 2 y d x d y x z d y d z ω = x 2 y d x d y x z d y d z omega=x^(2)ydx^^dy-xzdy^^dz\omega=x^{2} y d x \wedge d y-x z d y \wedge d zω=x2ydxdyxzdydz is a differential 2-form on
3 3 ^(3){ }^{3}3. On the space T ( 1 , 2 , 3 ) 3 T ( 1 , 2 , 3 ) 3 T_((1,2,3))^(3)T_{(1,2,3)}{ }^{3}T(1,2,3)3 it is just the 2-form 2 d x d y 3 d y d z 2 d x d y 3 d y d z 2dx^^dy-3dy^^dz2 d x \wedge d y-3 d y \wedge d z2dxdy3dydz. We will denote vectors in T ( 1 , 2 , 3 ) 3 T ( 1 , 2 , 3 ) 3 T_((1,2,3))^(3)T_{(1,2,3)}{ }^{3}T(1,2,3)3 as d x , d y , d z ( 1 , 2 , 3 ) d x , d y , d z ( 1 , 2 , 3 ) (:dx,dy,dz:)_((1,2,3))\langle d x, d y, d z\rangle_{(1,2,3)}dx,dy,dz(1,2,3). Hence, the value of ω ( 4 , 0 , 1 ( 1 , 2 , 3 ) , ( 3 , 1 , 2 ( 1 , 2 , 3 ) ) ω 4 , 0 , 1 ( 1 , 2 , 3 ) , ( 3 , 1 , 2 ( 1 , 2 , 3 ) omega((:4,0,-1:)_((1,2,3)),(3,1,2:)_((1,2,3)))\omega\left(\langle 4,0,-1\rangle_{(1,2,3)},(3,1,2\rangle_{(1,2,3)}\right)ω(4,0,1(1,2,3),(3,1,2(1,2,3)) is the same as the 2 -form, 2 d x d y + d y d z 2 d x d y + d y d z 2dx^^dy+dy^^dz2 d x \wedge d y+d y \wedge d z2dxdy+dydz, evaluated on the vectors 4 , 0 , 1 4 , 0 , 1 (:4,0,-1:)\langle 4,0,-1\rangle4,0,1 and 3 , 1 , 2 3 , 1 , 2 (:3,1,2:)\langle 3,1,2\rangle3,1,2, which we compute:
ω ( 4 , 0 , 1 ( 1 , 2 , 3 ) , 3 , 1 , 2 ( 1 , 2 , 3 ) ) = 2 d x d y 3 d y d z ( 4 , 0 , 1 , 3 , 1 , 2 ) = 2 | 4 3 0 1 | 3 | 0 1 1 2 | = 5 ω 4 , 0 , 1 ( 1 , 2 , 3 ) , 3 , 1 , 2 ( 1 , 2 , 3 ) = 2 d x d y 3 d y d z ( 4 , 0 , 1 , 3 , 1 , 2 ) = 2 4 3 0 1 3 0 1 1 2 = 5 {:[omega((:4,0,-1:)_((1,2,3)):}{:,(:3,1,2:)_((1,2,3)))],[=2dx^^dy-3dy^^dz((:4","0","-1:)","(:3","1","2:))],[=2|[4,3],[0,1]|-3|[0,1],[-1,2]|=5]:}\begin{aligned} \omega\left(\langle 4,0,-1\rangle_{(1,2,3)}\right. & \left.,\langle 3,1,2\rangle_{(1,2,3)}\right) \\ & =2 d x \wedge d y-3 d y \wedge d z(\langle 4,0,-1\rangle,\langle 3,1,2\rangle) \\ & =2\left|\begin{array}{ll} 4 & 3 \\ 0 & 1 \end{array}\right|-3\left|\begin{array}{rr} 0 & 1 \\ -1 & 2 \end{array}\right|=5 \end{aligned}ω(4,0,1(1,2,3),3,1,2(1,2,3))=2dxdy3dydz(4,0,1,3,1,2)=2|4301|3|0112|=5
Suppose ω ω omega\omegaω is a differential 2-form on . What does ω ω omega\omegaω act on? It takes a pair of vectors at each point of 3 3 ^(3){ }^{3}3 and returns a number. In other words, it takes two vector fields and returns a function from 3 3 ^(3){ }^{3}3 to . A vector field is simply a choice of vector in T p 3 T p 3 T_(p)^(3)T_{p}{ }^{3}Tp3, for each p p ppp R 3 R 3 inR^(3)\in \mathbb{R}^{3}R3. In general, a differential n n nnn-form on R m R m R^(m)\mathbb{R}^{m}Rm acts on n n nnn vector fields to produce a function from m m ^(m){ }^{m}m to (see Fig. 5.1).
Fig. 5.1. A differential 2-form, ω ω omega\omegaω, acts on a pair of vector fields, and returns a function from n n ^(n){ }^{n}n to
Example 19. V 1 = 2 y , 0 , x ( x , y , z ) V 1 = 2 y , 0 , x ( x , y , z ) V_(1)=(:2y,0,-x:)_((x,y,z))V_{1}=\langle 2 y, 0,-x\rangle_{(x, y, z)}V1=2y,0,x(x,y,z) is a vector field on R 3 R 3 R^(3)R^{3}R3. For example, it contains the vector 4 , 0 , 1 T ( 1 , 2 , 3 ) 3 4 , 0 , 1 T ( 1 , 2 , 3 ) 3 (:4,0,-1:)inT_((1,2,3))^(3)\langle 4,0,-1\rangle \in T_{(1,2,3)}{ }^{3}4,0,1T(1,2,3)3. If V 2 = z , 1 V 2 = z , 1 V_(2)=(:z,1V_{2}=\langle z, 1V2=z,1, x y ( x , y , z ) x y ( x , y , z ) xy:)_((x,y,z))x y\rangle_{(x, y, z)}xy(x,y,z) and ω ω omega\omegaω is the differential 2-form, x 2 y d x d y x z d y d z x 2 y d x d y x z d y d z x^(2)ydx^^dy-xzdy^^dzx^{2} y d x \wedge d y-x z d y \wedge d zx2ydxdyxzdydz, then which is a function from 3 3 ^(3){ }^{3}3 to
ω ( V 1 , V 2 ) = x 2 y d x d y x z d y d z ( 2 y , 0 , x ( x , y , z ) , z , 1 , x y ( x , y , z ) ) = x 2 y | 2 y z 0 1 | x z | 0 1 x x y | = 2 x 2 y 2 x 2 z ω V 1 , V 2 = x 2 y d x d y x z d y d z 2 y , 0 , x ( x , y , z ) , z , 1 , x y ( x , y , z ) = x 2 y 2 y z 0 1 x z 0 1 x x y = 2 x 2 y 2 x 2 z {:[omega(V_(1),V_(2))=x^(2)ydx^^dy-xzdy^^dz((:2y,0,x:)_((x,y,z)),(:z,1,xy:)_((x,y,z)))],[=x^(2)y|[2y,z],[0,1]|-xz|[0,1],[-x,xy]|=2x^(2)y^(2)-x^(2)z]:}\begin{aligned} \omega\left(V_{1}, V_{2}\right) & =x^{2} y d x \wedge d y-x z d y \wedge d z\left(\langle 2 y, 0, x\rangle_{(x, y, z)},\langle z, 1, x y\rangle_{(x, y, z)}\right) \\ & =x^{2} y\left|\begin{array}{rr} 2 y & z \\ 0 & 1 \end{array}\right|-x z\left|\begin{array}{rr} 0 & 1 \\ -x & x y \end{array}\right|=2 x^{2} y^{2}-x^{2} z \end{aligned}ω(V1,V2)=x2ydxdyxzdydz(2y,0,x(x,y,z),z,1,xy(x,y,z))=x2y|2yz01|xz|01xxy|=2x2y2x2z
Notice that V 2 V 2 V_(2)V_{2}V2 contains the vector 3 , 1 , 2 ( 1 , 2 , 3 ) 3 , 1 , 2 ( 1 , 2 , 3 ) (:3,1,2:)_((1,2,3))\langle 3,1,2\rangle_{(1,2,3)}3,1,2(1,2,3). So, from the previous example we would expect that 2 x 2 y 2 x 2 z 2 x 2 y 2 x 2 z 2x^(2)y^(2)-x^(2)z2 x^{2} y^{2}-x^{2} z2x2y2x2z equals 5 at the point ( 1 , 2 , 3 1 , 2 , 3 1,2,31,2,31,2,3 ), which is indeed the case.
5.1. Let ω ω omega\omegaω be the differential 2 -form on 3 3 ^(3){ }^{3}3 given by
ω = x y z d x d y + x 2 z d y d z y d x d z ω = x y z d x d y + x 2 z d y d z y d x d z omega=xyzdx^^dy+x^(2)zdy^^dz-ydx^^dz\omega=x y z d x \wedge d y+x^{2} z d y \wedge d z-y d x \wedge d zω=xyzdxdy+x2zdydzydxdz
Let V 1 V 1 V_(1)V_{1}V1 and V 2 V 2 V_(2)V_{2}V2 be the following vector fields:
V 1 = y , z , x 2 ( x , y , z ) , V 2 = x y , x z , y ( x , y , z ) . V 1 = y , z , x 2 ( x , y , z ) , V 2 = x y , x z , y ( x , y , z ) . V_(1)=(:y,z,x^(2):)_((x,y,z)),V_(2)=(:xy,xz,y:)_((x,y,z)).V_{1}=\left\langle y, z, x^{2}\right\rangle_{(x, y, z)}, V_{2}=\langle x y, x z, y\rangle_{(x, y, z)} .V1=y,z,x2(x,y,z),V2=xy,xz,y(x,y,z).
  1. What vectors do V 1 V 1 V_(1)V_{1}V1 and V 2 V 2 V_(2)V_{2}V2 contain at the point ( 1 , 2 , 3 ) ( 1 , 2 , 3 ) (1,2,3)(1,2,3)(1,2,3) ?
  2. Which 2-form is ω ω omega\omegaω on T ( 1 , 2 , 3 ) 3 T ( 1 , 2 , 3 ) 3 T_((1,2,3))^(3)T_{(1,2,3)}{ }^{3}T(1,2,3)3 ?
  3. Use your answers to the previous two questions to compute ω ( V 1 , V 2 ) ω V 1 , V 2 omega(V_(1),V_(2))\omega\left(V_{1}, V_{2}\right)ω(V1,V2) at the point ( 1 , 2 , 3 ) ( 1 , 2 , 3 ) (1,2,3)(1,2,3)(1,2,3).
  4. Compute ω ( V 1 , V 2 ) ω V 1 , V 2 omega(V_(1),V_(2))\omega\left(V_{1}, V_{2}\right)ω(V1,V2) at the point ( x , y , z ) ( x , y , z ) (x,y,z)(x, y, z)(x,y,z). Then plug in x = 1 x = 1 x=1x=1x=1, y = 2 y = 2 y=2y=2y=2 and z = 3 z = 3 z=3z=3z=3 to check your answer against the previous question.

5.2 Integrating differential 2-forms

Let's now recall the steps involved with integration of functions on subsets of 2 2 ^(2){ }^{2}2, which we learned in Section 1.3. Suppose R 2 R 2 R sub^(2)R \subset{ }^{2}R2 and f : R f : R f:R rarrf: R \rightarrowf:R. The following steps define the integral of f f fff over R R RRR :
  1. Choose a lattice of points in R , { ( x i , y j ) } R , x i , y j R,{(x_(i),y_(j))}R,\left\{\left(x_{i}, y_{j}\right)\right\}R,{(xi,yj)}.
  2. For each i , j i , j i,ji, ji,j define V i , j = ( x i + 1 , y j ) ( x i , y j ) V i , j = x i + 1 , y j x i , y j V_(i,j)=(x_(i+1),y_(j))-(x_(i),y_(j))V_{i, j}=\left(x_{i+1}, y_{j}\right)-\left(x_{i}, y_{j}\right)Vi,j=(xi+1,yj)(xi,yj) and V i , j = 2 = ( x i V i , j = 2 = x i V_(i,j=)^(2)=(x_(i):}V_{i, j=}^{2}=\left(x_{i}\right.Vi,j=2=(xi, y j + 1 ) ( x i , y j ) y j + 1 x i , y j {:y_(j+1))-(x_(i,)y_(j))\left.y_{j+1}\right)-\left(x_{i,} y_{j}\right)yj+1)(xi,yj) (See Fig. 5.2). Notice that V 1 i , j V 1 i , j V^(1)_(i,j)V^{1}{ }_{i, j}V1i,j and V 2 i , j V 2 i , j V^(2)_(i,j)V^{2}{ }_{i, j}V2i,j are both vectors in T ( x i , y j ) 2 T ( x i , y j ) 2 T_((xi,yj))^(2)T_{(x i, y j)}{ }^{2}T(xi,yj)2.
  3. For each i , j i , j i,ji, ji,j compute f ( x i , y j ) f x i , y j f(x_(i,)y_(j))f\left(x_{i,} y_{j}\right)f(xi,yj) Area ( V i , j r 1 V i , j 2 ) V i , j r 1 V i , j 2 (V_(i,jr)^(1)V_(i,j)^(2))\left(V_{i, j r}^{1} V_{i, j}^{2}\right)(Vi,jr1Vi,j2), where Area ( V ( V (V(V(V, W ) W ) W)W)W) is the function which returns the area of the parallelogram spanned by the vectors V V VVV and W W WWW.
  4. Sum over all i i iii and j j jjj.
  5. Take the limit as the maximal distance between adjacent lattice points goes to zero. This is the number that we define to be the value of R f d x d y R f d x d y int_(R)fdxdy\int_{R} f d x d yRfdxdy.
Let's focus on Step 3. Here we compute f ( x i , y j ) f x i , y j f(x_(i,)y_(j))f\left(x_{i,} y_{j}\right)f(xi,yj) Area ( V 1 i , j , V 2 i , j ) V 1 i , j , V 2 i , j (V^(1)_(i,j),V^(2)_(i,j))\left(V^{1}{ }_{i, j}, V^{2}{ }_{i, j}\right)(V1i,j,V2i,j). Notice that this is exactly the value of the differential 2 -form ω = f ω = f omega=f\omega=fω=f ( x , y ) d x d y ( x , y ) d x d y (x,y)dx^^dy(x, y) d x \wedge d y(x,y)dxdy, evaluated on the vectors V i , j 1 V i , j 1 V_(i,j)^(1)V_{i, j}^{1}Vi,j1 and V 2 i , j V 2 i , j V^(2)_(i,j)V^{2}{ }_{i, j}V2i,j at the point ( x i x i x_(i)x_{i}xi, y j ) y j {:y_(j))\left.y_{j}\right)yj). Hence, in Step 4 we can write this sum as , j f ( x i , y j ) , j f x i , y j sum,sum_(j)f(x_(i,)y_(j))\sum, \sum_{j} f\left(x_{i,} y_{j}\right),jf(xi,yj) Area ( V 1 i , j , V i , j 2 ) = i j ω ( x i , y j ) ( V 1 i , j , V i , j 2 ) V 1 i , j , V i , j 2 = i j ω x i , y j V 1 i , j , V i , j 2 (V^(1)_(i,j),V_(i,j)^(2))=sum i sum j omega(x_(i,)y_(j))(V^(1)_(i,j),V_(i,j)^(2))\left(V^{1}{ }_{i, j}, V_{i, j}^{2}\right)=\sum i \sum j \omega\left(x_{i,} y_{j}\right)\left(V^{1}{ }_{i, j}, V_{i, j}^{2}\right)(V1i,j,Vi,j2)=ijω(xi,yj)(V1i,j,Vi,j2). It is reasonable, then o dopt the shorthand " R ω R ω int_(R)omega\int_{R} \omegaRω " to denote the limit in Step 5.
The upshot of all this is the following:
If ω = f ( x , y ) d x d y , then R ω = R f d x d y .  If  ω = f ( x , y ) d x d y ,  then  R ω = R f d x d y . " If "omega=f(x,y)dx^^dy," then "int_(R)omega=int_(R)fdxdy.\text { If } \omega=f(x, y) d x \wedge d y, \text { then } \int_{R} \omega=\int_{R} f d x d y . If ω=f(x,y)dxdy, then Rω=Rfdxdy.
Fig. 5.2. The steps toward integration.
Since all differential 2-forms on 2 2 ^(2){ }^{2}2 are of the form f ( x , y ) d x d y f ( x , y ) d x d y f(x,y)dx^^dyf(x, y) d x \wedge d yf(x,y)dxdy we now know how to integrate them.
CAUTION! When integrating 2 -forms on 2 it is tempting to always drop the " Λ Λ Lambda\LambdaΛ " and forget you have a differential form. This is only valid with d x d y d x d y dx^^dyd x \wedge d ydxdy. It is NOT valid with d y d x d y d x dy^^dxd y \wedge d xdydx. This may seem a bit curious since Fubini's theorem gives us
f d x d y = f d x d y = f d y d x . f d x d y = f d x d y = f d y d x . int fdx^^dy=int fdxdy=int fdydx.\int f d x \wedge d y=\int f d x d y=\int f d y d x .fdxdy=fdxdy=fdydx.
All of these are equal to - f d y d x f d y d x int fdy^^dx\int f d y \wedge d xfdydx. We will revisit this issue in Example 27.
5.2. Let ω = x y 2 d x d y ω = x y 2 d x d y omega=xy^(2)dx^^dy\omega=x y^{2} d x \wedge d yω=xy2dxdy e a differential 2-form on 2. Let D D DDD be the region of 2 2 ^(2){ }^{2}2 where 0 x 1 0 x 1 0 <= x <= 10 \leq x \leq 10x1 and 0 y 1 0 y 1 0 <= y <= 10 \leq y \leq 10y1. Calculate D ω D ω int_(D)omega\int_{D} \omegaDω.
What about integration of differential 2-forms on 3 3 ^(3){ }^{3}3 ? As remarked at the end of Section 3.5 we do this only over those subsets of 3 3 ^(3){ }^{3}3 which can be parameterized by subsets of 2 2 ^(2){ }^{2}2. Suppose M M MMM is such a subset, like the top half of the unit sphere. To define what we mean by ω ω int omega\int \omegaω we just follow the steps above: M M MMM
  1. Choose a lattice of points in M , { p i , j } M , p i , j M_(,){p_(i,j)}M_{,}\left\{p_{i, j}\right\}M,{pi,j}.
  2. For each i , j i , j i,ji, ji,j define V i , j 1 = p i + 1 , j p i , j V i , j 1 = p i + 1 , j p i , j V_(i,j)^(1)=p_(i+1,j)-p_(i,j)V_{i, j}^{1}=p_{i+1, j}-p_{i, j}Vi,j1=pi+1,jpi,j and V i , j 2 = p i , j + 1 p i , j V i , j 2 = p i , j + 1 p i , j V_(i,j)^(2)=p_(i,j+1)-p_(i,j)V_{i, j}^{2}=p_{i, j+1}-p_{i, j}Vi,j2=pi,j+1pi,j. Notice that V i , j 1 V i , j 1 V_(i,j)^(1)V_{i, j}^{1}Vi,j1 and V i , j 2 V i , j 2 V_(i,j)^(2)V_{i, j}^{2}Vi,j2 are both vectors in T p i , j 3 T p i , j 3 T_(pi,j)^(3)T_{p i, j}{ }^{3}Tpi,j3 (see Fig. 5.3).
  3. For each i , j i , j i,ji, ji,j compute ω p i , j ( V i , j 1 V i , j 2 ) ω p i , j V i , j 1 V i , j 2 omegap_(i,j)(V_(i,j)^(1)V_(i,j)^(2))\omega p_{i, j}\left(V_{i, j}^{1} V_{i, j}^{2}\right)ωpi,j(Vi,j1Vi,j2).
  4. Sum over all i i iii and j j jjj.
  5. Take the limit as the maximal distance between adjacent lattice points goes to 0 . This is the number that we define to be the value of ω ω int omega\int \omegaω. M M MMM
Fig. 5.3. The steps toward integrating a 2 -form.
Unfortunately these steps are not so easy to follow. For one thing, it is not always clear how to pick the lattice in Step 1. In fact, there is an even worse prob lem. In Step 3, why did we compute ω p i , j ω p i , j omegap_(i,j)\omega p_{i, j}ωpi,j ( V i , j 1 V i , j 2 ) V i , j 1 V i , j 2 (V_(i,j)^(1)V_(i,j)^(2))\left(V_{i, j}^{1} V_{i, j}^{2}\right)(Vi,j1Vi,j2) instead of ω p i , j ( V i , j 2 V i , j 1 ) ω p i , j V i , j 2 V i , j 1 omegap_(i,j)(V_(i,j)^(2)V_(i,j)^(1))\omega p_{i, j}\left(V_{i, j}^{2} V_{i, j}^{1}\right)ωpi,j(Vi,j2Vi,j1) ? After all, V i , j 1 V i , j 1 V_(i,j)^(1)V_{i, j}^{1}Vi,j1 and V i , j 2 V i , j 2 V_(i,j)^(2)V_{i, j}^{2}Vi,j2 are two randomly oriented vectors in 7 3 p i , j 7 3 p i , j 7^(3)_(pi,j)7^{3}{ }_{p i, j}73pi,j. There is no reasonable way to decide which should be first and which second. There is nothing to be done about this. At some point we just have to make a choice and make it clear which choice we have made. Such a decision is called an orientation. We will have much more to say about this later. For now, we simply note that a different choice will only change our answer by changing its sign.
While we are on this topic, we also note that we would end up with the same number in Step 5 if we had calculated ω p i , j ( V 1 i , j ω p i , j V 1 i , j omegap_(i,j)(-V^(1)_(i,j)-:}\omega p_{i, j}\left(-V^{1}{ }_{i, j}-\right.ωpi,j(V1i,j V i , j 2 V i , j 2 V_(i,j)^(2)V_{i, j}^{2}Vi,j2 ) in Step 4, instead. Similarly, if it turns out later that we should have calculated ω p i , j ( V i , j 2 , V i , j 1 ) ω p i , j V i , j 2 , V i , j 1 omega_(pi,j)(V_(i,j)^(2),V_(i,j)^(1))\omega_{p i, j}\left(V_{i, j}^{2}, V_{i, j}^{1}\right)ωpi,j(Vi,j2,Vi,j1), then we could have also arrived at the right answer by computing ω p i , j ( V 1 i , j V 2 i , j ) ω p i , j V 1 i , j V 2 i , j omega_(pi,j)(-V^(1)_(i,j)V^(2)_(i,j))\omega_{p i, j}\left(-V^{1}{ }_{i, j} V^{2}{ }_{i, j}\right)ωpi,j(V1i,jV2i,j). In other words, there are really only two possibilities: either ω p i , j ( V i , j 1 V i , j 2 ) ω p i , j V i , j 1 V i , j 2 omega_(pi,j)(V_(i,j)^(1)V_(i,j)^(2))\omega_{p i, j}\left(V_{i, j}^{1} V_{i, j}^{2}\right)ωpi,j(Vi,j1Vi,j2) gives the correct answer or ω pi,j ( V i , j 1 , V i , j 2 ) ω pi,j  V i , j 1 , V i , j 2 omega_("pi,j ")(-V_(i,j)^(1),V_(i,j)^(2))\omega_{\text {pi,j }}\left(-V_{i, j}^{1}, V_{i, j}^{2}\right)ωpi,j (Vi,j1,Vi,j2) does. Which one will depend on our choice of orientation.
Despite all the difficulties with using the above definition of ω ω int omega\int \omegaω, all hope M M MMM is not lost. Remember that we are only integrating over regions which can be parameterized by subsets of R 2 R 2 R^(2)R^{2}R2. The trick is to use such a parameterization to translate the problem into an integral of a 2 -form over a region in R 2 R 2 R^(2)\mathrm{R}^{2}R2. The steps are analogous to those in Section 3.5.
Suppose φ : R R 2 M φ : R R 2 M varphi:R subR^(2)rarr M\varphi: R \subset R^{2} \rightarrow Mφ:RR2M is a parameterization. We want to find a 2-form, f ( x , y ) d x d y f ( x , y ) d x d y f(x,y)dx^^dyf(x, y) d x \wedge d yf(x,y)dxdy, such that a Riemann sum for this 2 -form over R R RRR gives the same result as a Riemann sum for ω ω omega\omegaω over M M MMM. Let's begin:
  1. Choose a rectangular lattice of points in R , { ( x i , y j ) } R , x i , y j R,{(x_(i),y_(j))}R,\left\{\left(x_{i}, y_{j}\right)\right\}R,{(xi,yj)}. This also gives a lattice, { Φ ( x i , y j ) } Φ x i , y j {Phi(x_(i),y_(j))}\left\{\Phi\left(x_{i}, y_{j}\right)\right\}{Φ(xi,yj)}, in M M MMM.
  2. For each i , j i , j i,ji, ji,j, define V 1 i , j = ( x i + 1 , y j ) ( x i , y j ) , V 2 i , j = ( x i , y j + 1 ) V 1 i , j = x i + 1 , y j x i , y j , V 2 i , j = x i , y j + 1 V^(1)_(i,j)=(x_(i+1),y_(j))-(x_(i,)y_(j)),V^(2)_(i,j)=(x_(i,)y_(j+1))-V^{1}{ }_{i, j}=\left(x_{i+1}, y_{j}\right)-\left(x_{i,} y_{j}\right), V^{2}{ }_{i, j}=\left(x_{i,} y_{j+1}\right)-V1i,j=(xi+1,yj)(xi,yj),V2i,j=(xi,yj+1) ( x i , y j ) , v 1 i , j = φ ( x i + 1 , y j ) φ ( x i , y j ) x i , y j , v 1 i , j = φ x i + 1 , y j φ x i , y j (x_(i),y_(j)),v^(1)_(i,j)=varphi(x_(i+1),y_(j))-varphi(x_(i),y_(j))\left(x_{i}, y_{j}\right), \mathrm{v}^{1}{ }_{i, j}=\varphi\left(x_{i+1}, y_{j}\right)-\varphi\left(x_{i}, y_{j}\right)(xi,yj),v1i,j=φ(xi+1,yj)φ(xi,yj), and v 2 i , j = φ ( x i , y j + 1 ) φ v 2 i , j = φ x i , y j + 1 φ v^(2)_(i,j)=varphi(x_(i),y_(j+1))-varphi\mathrm{v}^{2}{ }_{i, j}=\varphi\left(x_{i}, y_{j+1}\right)-\varphiv2i,j=φ(xi,yj+1)φ ( x i , y j ) x i , y j (x_(i,)y_(j))\left(x_{i,} y_{j}\right)(xi,yj) (see Fig. 5.4). Notice that V 1 i , j V 1 i , j V^(1)_(i,j)V^{1}{ }_{i, j}V1i,j and V 2 i , j V 2 i , j V^(2)_(i,j)V^{2}{ }_{i, j}V2i,j are vectors in T ( x i , y j ) R 2 T ( x i , y j ) R 2 T_((xi,yj))R^(2)T_{(x i, y j)} \mathrm{R}^{2}T(xi,yj)R2 and v 1 i , j v 1 i , j v^(1)_(i,j)\mathrm{v}^{1}{ }_{i, j}v1i,j and v 2 i , j v 2 i , j v^(2)_(i,j)\mathrm{v}^{2}{ }_{i, j}v2i,j are vectors in T φ ( x i , y ) R 3 T φ ( x i , y ) R 3 Tvarphi_((xi,y))R^(3)T \varphi_{(x i, y)} \mathrm{R}^{3}Tφ(xi,y)R3.
  3. For each i , j i , j i,ji, ji,j compute f ( x i , y j ) d x d y ( V i , j , 1 V i , j 2 ) f x i , y j d x d y V i , j , 1 V i , j 2 f(x_(i,)y_(j))dx^^dy(V_(i,j,)^(1)V_(i,j)^(2))f\left(x_{i,} y_{j}\right) d x \wedge d y\left(V_{i, j,}^{1} V_{i, j}^{2}\right)f(xi,yj)dxdy(Vi,j,1Vi,j2) and ω φ ( x i , y j ) ω φ ( x i , y j ) omegavarphi_((xi,yj))\omega \varphi_{(x i, y j)}ωφ(xi,yj) ( V 1 i , j , v 2 i , j ) V 1 i , j , v 2 i , j (V^(1)_(i,j),v^(2)_(i,j))\left(\mathrm{V}^{1}{ }_{i, j}, \mathrm{v}^{2}{ }_{i, j}\right)(V1i,j,v2i,j).
  4. Sum over all i i iii and j j jjj.
Fig. 5.4. Using φ φ varphi\varphiφ to integrate a 2 -form.
At the conclusion of Step 4 we have two sums, Σ i Σ j f ( X i , y j ) d x Σ i Σ j f X i , y j d x Sigma_(i)Sigma_(j)f(X_(i),y_(j))dx^^\Sigma_{i} \Sigma_{j} f\left(X_{i}, y_{j}\right) d x \wedgeΣiΣjf(Xi,yj)dx d y ( V 1 i , j V 2 i , j ) d y V 1 i , j V 2 i , j dy(V^(1)_(i,j)V^(2)_(i,j))d y\left(V^{1}{ }_{i, j} V^{2}{ }_{i, j}\right)dy(V1i,jV2i,j) and Σ i Σ j ω φ ( x i , y j ) ( V 1 i , j v 2 i , j ) Σ i Σ j ω φ ( x i , y j ) V 1 i , j v 2 i , j Sigma_(i)Sigma_(j)omega varphi(xi,yj)(V^(1)_(i,j)v^(2)_(i,j))\Sigma_{i} \Sigma_{j} \omega \varphi(x i, y j)\left(V^{1}{ }_{i, j} v^{2}{ }_{i, j}\right)ΣiΣjωφ(xi,yj)(V1i,jv2i,j). In order for these to be equal, we must have:
f ( x i , y j ) d x d y ( V i , j 1 , V i , j 2 ) = ω ϕ ( x i , y j ) ( V i , j 1 , V i , j 2 ) f x i , y j d x d y V i , j 1 , V i , j 2 = ω ϕ x i , y j V i , j 1 , V i , j 2 f(x_(i),y_(j))dx^^dy(V_(i,j)^(1),V_(i,j)^(2))=omega_(phi(x_(i),y_(j)))(V_(i,j)^(1),V_(i,j)^(2))f\left(x_{i}, y_{j}\right) d x \wedge d y\left(V_{i, j}^{1}, V_{i, j}^{2}\right)=\omega_{\phi\left(x_{i}, y_{j}\right)}\left(\mathcal{V}_{i, j}^{1}, V_{i, j}^{2}\right)f(xi,yj)dxdy(Vi,j1,Vi,j2)=ωϕ(xi,yj)(Vi,j1,Vi,j2)
And so,
f ( x i , y j ) = ω ϕ ( x i , y j ) ( V i , j 1 , ν i , j 2 ) d x d y ( V i , j 1 , V i , j 2 ) . f x i , y j = ω ϕ x i , y j V i , j 1 , ν i , j 2 d x d y V i , j 1 , V i , j 2 . f(x_(i),y_(j))=(omega_(phi(x_(i),y_(j)))(V_(i,j)^(1),nu_(i,j)^(2)))/(dx^^dy(V_(i,j)^(1),V_(i,j)^(2))).f\left(x_{i}, y_{j}\right)=\frac{\omega_{\phi\left(x_{i}, y_{j}\right)}\left(\mathcal{V}_{i, j}^{1}, \nu_{i, j}^{2}\right)}{d x \wedge d y\left(V_{i, j}^{1}, V_{i, j}^{2}\right)} .f(xi,yj)=ωϕ(xi,yj)(Vi,j1,νi,j2)dxdy(Vi,j1,Vi,j2).
But, since we are using a rectangular lattice in R R RRR we know d x d y d x d y dx^^dyd x \wedge d ydxdy ( V i , j 1 V i , j 2 ) = Area ( V 1 i , j V i , j 2 ) = | V i , j 1 | | V 2 i , j | V i , j 1 V i , j 2 = Area V 1 i , j V i , j 2 = V i , j 1 V 2 i , j (V_(i,j)^(1)V_(i,j)^(2))=Area(V^(1)_(i,j)V_(i,j)^(2))=|V_(i,j)^(1)|*|V^(2)_(i,j)|\left(V_{i, j}^{1} V_{i, j}^{2}\right)=\operatorname{Area}\left(V^{1}{ }_{i, j} V_{i, j}^{2}\right)=\left|V_{i, j}^{1}\right| \cdot\left|V^{2}{ }_{i, j}\right|(Vi,j1Vi,j2)=Area(V1i,jVi,j2)=|Vi,j1||V2i,j|. We now have
f ( x i , y j ) = ω ϕ ϕ ( x i , j ) ( v i , j 1 , v i , j 2 ) | V i , j 1 | | | V i , j 2 . f x i , y j = ω ϕ ϕ x i , j v i , j 1 , v i , j 2 V i , j 1 | | V i , j 2 . f(x_(i),y_(j))=(omega_(phi phi(x_(i,j)))(v_(i,j)^(1),v_(i,j)^(2)))/(|V_(i,j)^(1)|*|*|V_(i,j)^(2)∣).f\left(x_{i}, y_{j}\right)=\frac{\omega_{\phi \phi\left(x_{i, j}\right)}\left(v_{i, j}^{1}, v_{i, j}^{2}\right)}{\left|V_{i, j}^{1}\right| \cdot|\cdot| V_{i, j}^{2} \mid} .f(xi,yj)=ωϕϕ(xi,j)(vi,j1,vi,j2)|Vi,j1|||Vi,j2.
Using the bilinearity of ω ω omega\omegaω this reduces to
f ( x i , y j ) = ω ϕ ( x i , y j ( v 1 1 , j 1 V i , j 1 , , v i , j 2 V i , j 2 ) . f x i , y j = ω ϕ x i , y j v 1 1 , j 1 V i , j 1 , , v i , j 2 V i , j 2 . f(x_(i),y_(j))=omega_(phi(x_(i),y_(j):})((v_((1)/(1),j)^(1))/(∣V_(i,j)^(1),),(v_(i,j)^(2))/(V_(i,j)^(2)∣)).f\left(x_{i}, y_{j}\right)=\omega_{\phi\left(x_{i}, y_{j}\right.}\left(\frac{v_{\frac{1}{1}, j}^{1}}{\mid V_{i, j}^{1},}, \frac{v_{i, j}^{2}}{V_{i, j}^{2} \mid}\right) .f(xi,yj)=ωϕ(xi,yj(v11,j1Vi,j1,,vi,j2Vi,j2).
But, as the distance between adjacent points of our partition tends toward 0,
V i , j 1 | V i , j 1 | = ϕ ( x i + 1 , y j ) ϕ ( x i , y j ) | ( x i + 1 , y j ) ( x i , y j ) | = ϕ ( x i + 1 , y j ) ϕ ( x i , y j ) | x i + 1 x i | ϕ x ( x i , y j ) . V i , j 1 V i , j 1 = ϕ x i + 1 , y j ϕ x i , y j x i + 1 , y j x i , y j = ϕ x i + 1 , y j ϕ x i , y j x i + 1 x i ϕ x x i , y j . (V_(i,j)^(1))/(|V_(i,j)^(1)|)=(phi(x_(i+1),y_(j))-phi(x_(i),y_(j)))/(|(x_(i+1),y_(j))-(x_(i),y_(j))|)=(phi(x_(i+1),y_(j))-phi(x_(i),y_(j)))/(|x_(i+1)-x_(i)|)rarr(del phi)/(del x)(x_(i),y_(j)).\frac{\mathcal{V}_{i, j}^{1}}{\left|V_{i, j}^{1}\right|}=\frac{\phi\left(x_{i+1}, y_{j}\right)-\phi\left(x_{i}, y_{j}\right)}{\left|\left(x_{i+1}, y_{j}\right)-\left(x_{i}, y_{j}\right)\right|}=\frac{\phi\left(x_{i+1}, y_{j}\right)-\phi\left(x_{i}, y_{j}\right)}{\left|x_{i+1}-x_{i}\right|} \rightarrow \frac{\partial \phi}{\partial x}\left(x_{i}, y_{j}\right) .Vi,j1|Vi,j1|=ϕ(xi+1,yj)ϕ(xi,yj)|(xi+1,yj)(xi,yj)|=ϕ(xi+1,yj)ϕ(xi,yj)|xi+1xi|ϕx(xi,yj).

Let's summarize what we have so far. We defined f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) so that
i j ω ϕ ( x i , y j ) ( V i , j 1 , V i , j 2 ) = i j f ( x i , y j ) d x d y ( V i , j 1 , V i , j 2 ) = i j ω ϕ ( x i , y j ) ( ν i , j 1 | V i , j 1 | , V i , j 2 | V i , j 2 | ) d x d y ( V i , j 1 , V i , j 2 ) . i j ω ϕ x i , y j V i , j 1 , V i , j 2 = i j f x i , y j d x d y V i , j 1 , V i , j 2 = i j ω ϕ x i , y j ν i , j 1 V i , j 1 , V i , j 2 V i , j 2 d x d y V i , j 1 , V i , j 2 . {:[sum_(i)sum_(j)omega_(phi(x_(i),y_(j)))(V_(i,j)^(1),V_(i,j)^(2))=sum_(i)sum_(j)f(x_(i),y_(j))dx^^dy(V_(i,j)^(1),V_(i,j)^(2))],[=sum_(i)sum_(j)omega_(phi(x_(i),y_(j)))((nu_(i,j)^(1))/(|V_(i,j)^(1)|),(V_(i,j)^(2))/(|V_(i,j)^(2)|))dx^^dy(V_(i,j)^(1),V_(i,j)^(2)).]:}\begin{gathered} \sum_{i} \sum_{j} \omega_{\phi\left(x_{i}, y_{j}\right)}\left(\mathcal{V}_{i, j}^{1}, V_{i, j}^{2}\right)=\sum_{i} \sum_{j} f\left(x_{i}, y_{j}\right) d x \wedge d y\left(V_{i, j}^{1}, V_{i, j}^{2}\right) \\ =\sum_{i} \sum_{j} \omega_{\phi\left(x_{i}, y_{j}\right)}\left(\frac{\nu_{i, j}^{1}}{\left|V_{i, j}^{1}\right|}, \frac{V_{i, j}^{2}}{\left|V_{i, j}^{2}\right|}\right) d x \wedge d y\left(V_{i, j}^{1}, V_{i, j}^{2}\right) . \end{gathered}ijωϕ(xi,yj)(Vi,j1,Vi,j2)=ijf(xi,yj)dxdy(Vi,j1,Vi,j2)=ijωϕ(xi,yj)(νi,j1|Vi,j1|,Vi,j2|Vi,j2|)dxdy(Vi,j1,Vi,j2).
We have also shown that when we take the limit as the distance between adjacent partition points tends toward zero this sum
converges to the sum
i j ω ϕ ( x , y ) ( ϕ x ( x , y ) , ϕ y ( x , y ) ) d x d y ( V i , j 1 , V i , j 2 ) . i j ω ϕ ( x , y ) ϕ x ( x , y ) , ϕ y ( x , y ) d x d y V i , j 1 , V i , j 2 . sum_(i)sum_(j)omega_(phi(x,y))((del phi)/(del x)(x,y),(del phi)/(del y)(x,y))dx^^dy(V_(i,j)^(1),V_(i,j)^(2)).\sum_{i} \sum_{j} \omega_{\phi(x, y)}\left(\frac{\partial \phi}{\partial x}(x, y), \frac{\partial \phi}{\partial y}(x, y)\right) d x \wedge d y\left(V_{i, j}^{1}, V_{i, j}^{2}\right) .ijωϕ(x,y)(ϕx(x,y),ϕy(x,y))dxdy(Vi,j1,Vi,j2).
Hence, it must be that
(5.1) M ω = R ω ϕ ( x , y ) ( ϕ x ( x , y ) , ϕ y ( x , y ) ) d x d y . (5.1) M ω = R ω ϕ ( x , y ) ϕ x ( x , y ) , ϕ y ( x , y ) d x d y . {:(5.1)int_(M)omega=int_(R)omega_(phi(x,y))((del phi)/(del x)(x,y),(del phi)/(del y)(x,y))dx^^dy.:}\begin{equation*} \int_{M} \omega=\int_{R} \omega_{\phi(x, y)}\left(\frac{\partial \phi}{\partial x}(x, y), \frac{\partial \phi}{\partial y}(x, y)\right) d x \wedge d y . \tag{5.1} \end{equation*}(5.1)Mω=Rωϕ(x,y)(ϕx(x,y),ϕy(x,y))dxdy.
At first glance, this seems like a very complicated formula. Let's break it down by examining the integrand on the right. The most important thing to notice is that this is just a differential 2 -form on R R RRR, even though ω ω omega\omegaω is a 2 -form on R 3 R 3 R^(3)\mathrm{R}^{3}R3. For each pair of numbers, ( x ( x (x(x(x, y y yyy ), the function ω ϕ ( x , y ) ( ϕ x ( x , y ) , ϕ y ( x , y ) ) just returns some real ω ϕ ( x , y ) ϕ x ( x , y ) , ϕ y ( x , y ) just   returns some real  ^(omega_(phi(x,y))((del phi)/(del x)(x,y),(del phi)/(del y)(x,y))_("just ")" returns some real "){ }^{\omega_{\phi(x, y)}\left(\frac{\partial \phi}{\partial x}(x, y), \frac{\partial \phi}{\partial y}(x, y)\right)_{\text {just }} \text { returns some real }}ωϕ(x,y)(ϕx(x,y),ϕy(x,y))just  returns some real  number. Hence, the entire integrand s of the form g d x d y g d x d y gdx^^dyg d x \wedge d ygdxdy, where g : R R g : R R g:R rarrRg: R \rightarrow \mathrm{R}g:RR.
The only way to really convince oneself of the usefulness of this formula is to actually use it.
Example 20. Let M M MMM denote the top half of the unit sphere in R 3 R 3 R^(3)\mathrm{R}^{3}R3. Let ω = z 2 d x d y ω = z 2 d x d y omega=z^(2)dx^^dy\omega=z^{2} d x \wedge d yω=z2dxdy be a differential 2-form on R 3 R 3 R^(3)\mathrm{R}^{3}R3. Calculating M ω M ω int_(M)omega\int_{M} \omegaMω directly by setting up a Riemann sum would be next to impossible. So we employ the parameterization Φ ( r , t ) = ( r cos t , r sin t , 1 r 2 ) Φ ( r , t ) = r cos t , r sin t , 1 r 2 Phi(r,t)=(r cos t,r sin t,sqrt()1-r^(2))\Phi(r, t)=\left(r \cos t, r \sin t, \sqrt{ } 1-r^{2}\right)Φ(r,t)=(rcost,rsint,1r2), where 0 t 2 π 0 t 2 π 0 <= t <= 2pi0 \leq t \leq 2 \pi0t2π and 0 r 1 0 r 1 0 <= r <= 10 \leq r \leq 10r1.
M ω = R ω ϕ ( r , t ) ( ϕ r ( r , t ) , ϕ t ( r , t ) ) d r d t = R ω ϕ ( r , t ) ( cos t , sin t , r 1 r 2 , r sin t , r cos t , 0 ) d r d t = R ( 1 r 2 ) | cos t r sin t sin t r cos t | d r d t = R ( 1 r 2 ) ( r ) d r d t = 0 2 π 0 1 r r 3 d r d t = π 2 . M ω = R ω ϕ ( r , t ) ϕ r ( r , t ) , ϕ t ( r , t ) d r d t = R ω ϕ ( r , t ) cos t , sin t , r 1 r 2 , r sin t , r cos t , 0 d r d t = R 1 r 2 cos t r sin t sin t r cos t d r d t = R 1 r 2 ( r ) d r d t = 0 2 π 0 1 r r 3 d r d t = π 2 . {:[int_(M)omega=int_(R)omega_(phi(r,t))((del phi)/(del r)(r,t),(del phi)/(del t)(r,t))dr^^dt],[=int_(R)omega_(phi(r,t))((:cos t,sin t,(-r)/(sqrt(1-r^(2))):),(:-r sin t,r cos t,0:))dr^^dt],[=int_(R)(1-r^(2))|[cos t-r sin t],[sin t quad r cos t]|dr^^dt],[=int_(R)(1-r^(2))(r)dr^^dt],[=int_(0)^(2pi)int_(0)^(1)r-r^(3)drdt=(pi)/(2).]:}\begin{aligned} \int_{M} \omega & =\int_{R} \omega_{\phi(r, t)}\left(\frac{\partial \phi}{\partial r}(r, t), \frac{\partial \phi}{\partial t}(r, t)\right) d r \wedge d t \\ & =\int_{R} \omega_{\phi(r, t)}\left(\left\langle\cos t, \sin t, \frac{-r}{\sqrt{1-r^{2}}}\right\rangle,\langle-r \sin t, r \cos t, 0\rangle\right) d r \wedge d t \\ & =\int_{R}\left(1-r^{2}\right)\left|\begin{array}{cc} \cos t-r \sin t \\ \sin t \quad r \cos t \end{array}\right| d r \wedge d t \\ & =\int_{R}\left(1-r^{2}\right)(r) d r \wedge d t \\ & =\int_{0}^{2 \pi} \int_{0}^{1} r-r^{3} d r d t=\frac{\pi}{2} . \end{aligned}Mω=Rωϕ(r,t)(ϕr(r,t),ϕt(r,t))drdt=Rωϕ(r,t)(cost,sint,r1r2,rsint,rcost,0)drdt=R(1r2)|costrsintsintrcost|drdt=R(1r2)(r)drdt=02π01rr3drdt=π2.
Notice that as promised, the term ω ϕ ( r , t ) ( ϕ r ( r , t ) , ϕ t ( r , t ) ) ω ϕ ( r , t ) ϕ r ( r , t ) , ϕ t ( r , t ) ^(omega_(phi(r,t)))((del phi)/(del r)(r,t),(del phi)/(del t)(r,t)){ }^{\omega_{\phi(r, t)}}\left(\frac{\partial \phi}{\partial r}(r, t), \frac{\partial \phi}{\partial t}(r, t)\right)ωϕ(r,t)(ϕr(r,t),ϕt(r,t)) in the second integral above simplified to a function from R R , r r 3 R R , r r 3 R@R,r-r^(3)R \circ R, r-r^{3}RR,rr3.
5.3. Integrate the 2 -form
ω = 1 x d y d z 1 y d x d z ω = 1 x d y d z 1 y d x d z omega=(1)/(x)dy^^dz-(1)/(y)dx^^dz\omega=\frac{1}{x} d y \wedge d z-\frac{1}{y} d x \wedge d zω=1xdydz1ydxdz
over the top half of the unit sphere using the following parameterizations from cylindrical and spherical coordinates:
  1. ( r , θ ) ( r cos θ , r sin θ , 1 r 2 ) ( r , θ ) r cos θ , r sin θ , 1 r 2 (r,theta)rarr(r cos theta,r sin theta,sqrt()1-r^(2))(r, \theta) \rightarrow\left(r \cos \theta, r \sin \theta, \sqrt{ } 1-r^{2}\right)(r,θ)(rcosθ,rsinθ,1r2), where 0 θ 2 π 0 θ 2 π 0 <= theta <= 2pi0 \leq \theta \leq 2 \pi0θ2π and 0 r 0 r 0 <= r0 \leq r0r 1 1 <= 1\leq 11.
  2. ( θ , φ ) ( sin φ cos θ , sin φ sin θ , cos φ ) ( θ , φ ) ( sin φ cos θ , sin φ sin θ , cos φ ) (theta,varphi)rarr(sin varphi cos theta,sin varphi sin theta,cos varphi)(\theta, \varphi) \rightarrow(\sin \varphi \cos \theta, \sin \varphi \sin \theta, \cos \varphi)(θ,φ)(sinφcosθ,sinφsinθ,cosφ), where 0 θ 2 π 0 θ 2 π 0 <= theta <= 2pi0 \leq \theta \leq 2 \pi0θ2π and 0 ϕ π 2 0 ϕ π 2 0 <= phi <= (pi)/(2)0 \leq \phi \leq \frac{\pi}{2}0ϕπ2
    5.4. Let ω ω omega\omegaω be the 2 -form from the previous problem. Integrate ω ω omega\omegaω over the surface parameterized by the following:
ϕ ( r , θ ) = ( r cos θ , r sin θ , cos r ) , 0 r π 2 , 0 θ 2 π . ϕ ( r , θ ) = ( r cos θ , r sin θ , cos r ) , 0 r π 2 , 0 θ 2 π . phi(r,theta)=(r cos theta,r sin theta,cos r),0 <= r <= (pi)/(2),0 <= theta <= 2pi.\phi(r, \theta)=(r \cos \theta, r \sin \theta, \cos r), 0 \leq r \leq \frac{\pi}{2}, 0 \leq \theta \leq 2 \pi .ϕ(r,θ)=(rcosθ,rsinθ,cosr),0rπ2,0θ2π.
5.5. Let S S SSS be the surface in R 3 R 3 R^(3)\mathrm{R}^{3}R3 parameterized by
Ψ ( θ , z ) = ( cos θ , sin θ , z ) Ψ ( θ , z ) = ( cos θ , sin θ , z ) Psi(theta,z)=(cos theta,sin theta,z)\Psi(\theta, z)=(\cos \theta, \sin \theta, z)Ψ(θ,z)=(cosθ,sinθ,z)
where 0 θ π 0 θ π 0 <= theta <= pi0 \leq \theta \leq \pi0θπ and 0 z 1 0 z 1 0 <= z <= 10 \leq z \leq 10z1. Let ω = x y z d y d z ω = x y z d y d z omega=xyzdy^^dz\omega=x y z d y \wedge d zω=xyzdydz. Calculates ω ω int^(int)omega\int^{\int} \omegaω.
5.6. Let ω ω omega\omegaω be the differential 2 -form on R 3 R 3 R^(3)R^{3}R3 given by
ω = x y z d x d y + x 2 z d y d z y d x d z ω = x y z d x d y + x 2 z d y d z y d x d z omega=xyzdx^^dy+x^(2)zdy^^dz-ydx^^dz\omega=x y z d x \wedge d y+x^{2} z d y \wedge d z-y d x \wedge d zω=xyzdxdy+x2zdydzydxdz
  1. Let P P PPP be the portion of the plane 3 = 2 x + 3 y z 3 = 2 x + 3 y z 3=2x+3y-z3=2 x+3 y-z3=2x+3yz in R 3 R 3 R^(3)\mathrm{R}^{3}R3 that lies above the square { ( x , y ) 0 x 1 , 0 y 1 } { ( x , y ) 0 x 1 , 0 y 1 } {(x,y)∣0 <= x <= 1,0 <= y <= 1}\{(x, y) \mid 0 \leq x \leq 1,0 \leq y \leq 1\}{(x,y)0x1,0y1}. Calculate P ω P ω int_(P)omega\int_{P} \omegaPω.
  2. Let M M MMM be the portion of the graph of z = x 2 + y z = x 2 + y z=x^(2)+yz=x^{2}+yz=x2+y in R 3 R 3 R^(3)\mathrm{R}^{3}R3 that lies above the rectangle { ( x , y ) 0 x 1 , 0 y 2 } { ( x , y ) 0 x 1 , 0 y 2 } {(x,y)∣0 <= x <= 1,0 <= y <= 2}\{(x, y) \mid 0 \leq x \leq 1,0 \leq y \leq 2\}{(x,y)0x1,0y2}. Calculate M ω . M M ω . M int_(M)omega.M\int_{M} \omega . MMω.M
    5.7. Let D D DDD be some region in the x y x y xyx yxy-plane. Let M M MMM denote the portion of the graph of z = g ( x , y ) z = g ( x , y ) z=g(x,y)z=g(x, y)z=g(x,y) that lies above D D DDD.
  3. Let ω = f ( x , y ) d x d y ω = f ( x , y ) d x d y omega=f(x,y)dx^^dy\omega=f(x, y) d x \wedge d yω=f(x,y)dxdy be a differential 2-form on R 3 R 3 R^(3)\mathrm{R}^{3}R3. Show that
M ω = D f ( x , y ) d x d y . M ω = D f ( x , y ) d x d y . int_(M)omega=int_(D)f(x,y)dxdy.\int_{M} \omega=\int_{D} f(x, y) d x d y .Mω=Df(x,y)dxdy.
Notice the answer does not depend on the function g ( x , y ) g ( x , y ) g(x,y)g(x, y)g(x,y). 2. Now suppose ω = f ( x , y , z ) d x d y ω = f ( x , y , z ) d x d y omega=f(x,y,z)dx^^dy\omega=f(x, y, z) d x \wedge d yω=f(x,y,z)dxdy. Show that
M ω = D f ( x , y , g ( x , y ) ) d x d y M ω = D f ( x , y , g ( x , y ) ) d x d y int_(M)omega=int_(D)f(x,y,g(x,y))dxdy\int_{M} \omega=\int_{D} f(x, y, g(x, y)) d x d yMω=Df(x,y,g(x,y))dxdy
5.8. Let S S SSS be the surface obtained from the graph of z = f ( x ) = x 3 z = f ( x ) = x 3 z=f(x)=x^(3)z=f(x)=x^{3}z=f(x)=x3, where 0 x 1 0 x 1 0 <= x <= 10 \leq x \leq 10x1, by rotating around the z z zzz-axis. Integrate the 2 -form ω = y d x d z ω = y d x d z omega=ydx^^dz\omega=y d x \wedge d zω=ydxdz over S S SSS. (Hint. use cylindrical coordinates to parameterize S S SSS.)

5.3 Orientations

What would have happened in Example 20 if we had used the parameterization φ ( r , t ) = ( r cos t , r sin t , 1 r 2 ) φ ( r , t ) = r cos t , r sin t , 1 r 2 varphi^(')(r,t)=(-r cos t,r sin t,sqrt()1-r^(2))\varphi^{\prime}(r, t)=\left(-r \cos t, r \sin t, \sqrt{ } 1-r^{2}\right)φ(r,t)=(rcost,rsint,1r2) instead? We leave it to the reader to check that we end up with the answer π / 2 π / 2 -pi//2-\pi / 2π/2 rather than π / 2 π / 2 pi//2\pi / 2π/2. This is a problem. We defined ω ω int omega\int \omegaω before we started talking about parameterizations. Hence, the value M M MMM which we calculate for this integral should not depend on our choice of parameterization. So what happened?
To analyze this completely, we need to go back to the definition of ω ω int omega\int \omegaω from M M MMM the previous section. We noted at the time that a choice was made to calculate ω p i , j ( V i , j 1 V i , j 2 ) ω p i , j V i , j 1 V i , j 2 omegap_(i,j)(V_(i,j)^(1)V_(i,j)^(2))\omega p_{i, j}\left(V_{i, j}^{1} V_{i, j}^{2}\right)ωpi,j(Vi,j1Vi,j2) instead of ω p i , j ( V i , j 1 V i , j 2 ) ω p i , j V i , j 1 V i , j 2 omegap_(i,j)(-V_(i,j)^(1)V_(i,j)^(2))\omega p_{i, j}\left(-V_{i, j}^{1} V_{i, j}^{2}\right)ωpi,j(Vi,j1Vi,j2). But was this choice correct? The answer is a resounding maybe! We are actually missing enough information to tell. An orientation is precisely some piece of information about M M MMM which we can use to make the right choice. This way we can tell a friend what M M MMM is, what ω ω omega\omegaω is, and what the orientation on M M MMM is, and they are sure to get the same answer. Recall Equation 5.1:
M ω = R ω ϕ ( x , y ) ( ϕ x ( x , y ) , ϕ y ( x , y ) ) d x d y . M ω = R ω ϕ ( x , y ) ϕ x ( x , y ) , ϕ y ( x , y ) d x d y . int_(M)omega=int_(R)omega_(phi(x,y))((del phi)/(del x)(x,y),(del phi)/(del y)(x,y))dx^^dy.\int_{M} \omega=\int_{R} \omega_{\phi(x, y)}\left(\frac{\partial \phi}{\partial x}(x, y), \frac{\partial \phi}{\partial y}(x, y)\right) d x \wedge d y .Mω=Rωϕ(x,y)(ϕx(x,y),ϕy(x,y))dxdy.
Depending on the specified orientation of M M MMM, it may be incorrect to use Equation 5.1. Sometimes we may want to use:
M ω = R ω ϕ ( x , y ) ( ϕ x ( x , y ) , ϕ y ( x , y ) ) d x d y . M ω = R ω ϕ ( x , y ) ϕ x ( x , y ) , ϕ y ( x , y ) d x d y . int_(M)omega=int_(R)omega_(phi(x,y))(-(del phi)/(del x)(x,y),(del phi)/(del y)(x,y))dx^^dy.\int_{M} \omega=\int_{R} \omega_{\phi(x, y)}\left(-\frac{\partial \phi}{\partial x}(x, y), \frac{\partial \phi}{\partial y}(x, y)\right) d x \wedge d y .Mω=Rωϕ(x,y)(ϕx(x,y),ϕy(x,y))dxdy.
Both ω ω omega\omegaω and int\int are linear. This just means the negative sign in the integrand on the right can go all the way outside. Hence, we can write both this equation and Equation 5.1 as
(5.2) M ω = ± R ω ϕ ( x , y ) ( ϕ x ( x , y ) , ϕ y ( x , y ) ) d x d y . (5.2) M ω = ± R ω ϕ ( x , y ) ϕ x ( x , y ) , ϕ y ( x , y ) d x d y . {:(5.2)int_(M)omega=+-int_(R)omega_(phi(x,y))((del phi)/(del x)(x,y),(del phi)/(del y)(x,y))dx^^dy.:}\begin{equation*} \int_{M} \omega= \pm \int_{R} \omega_{\phi(x, y)}\left(\frac{\partial \phi}{\partial x}(x, y), \frac{\partial \phi}{\partial y}(x, y)\right) d x \wedge d y . \tag{5.2} \end{equation*}(5.2)Mω=±Rωϕ(x,y)(ϕx(x,y),ϕy(x,y))dxdy.
We define an orientation on M M MMM to be any piece of information that can be used to decide, for each choice of parameterization φ φ varphi\varphiφ, whether to use the " + " or "-" sign in Equation 5.2, so that the integral will always yield the same answer.
We will see several ways to specify an orientation on M M MMM. The first will be geometric. It has the advantage that it can be easily visualized, but the disadvantage that it is actually much harder to use in calculations. All we do is draw a small circle on M M MMM with an arrowhead on it. To use this "oriented circle" to tell if we need the " + " or "-" sign in Equation 5.2, we draw the vectors ϕ x ( x , y ) ϕ x ( x , y ) (del phi)/(del x)(x,y)\frac{\partial \phi}{\partial x}(x, y)ϕx(x,y) and ϕ y ( x , y ) ϕ y ( x , y ) (del phi)/(del y)(x,y)\frac{\partial \phi}{\partial y}(x, y)ϕy(x,y) and an arc with an arrow from the first to the second. If the direction of this arrow agrees with the oriented circle, then we use the " + " sign. If they disagree, then we use the "-" sign. See Figure 5.5.
Fig. 5.5. An orientation on M M MMM is given by an oriented circle.
Use the " - " sign when integrating.
Use the " + " sign when integrating.
A more algebraic way to specify an orientation is to simply pick a point p p ppp of M M MMM and choose any 2-form vv\vee on T p R 3 T p R 3 T_(p)R^(3)T_{p} \mathrm{R}^{3}TpR3 such that v ( V 1 p , V 2 p ) v V 1 p , V 2 p v(V^(1)_(p),V^(2)_(p))\mathrm{v}\left(V^{1}{ }_{p}, V^{2}{ }_{p}\right)v(V1p,V2p) 0 0 ✓!=0\checkmark \neq 00 whenever V p and 1 V p 2 V p  and  1 V p 2 V_(p" and ")^(1)V_(p)^(2)V_{p \text { and }}^{1} V_{p}^{2}Vp and 1Vp2 are vectors tangent to M M MMM, and V 1 V 1 V_(1)V_{1}V1 is not a multiple of V 2 V 2 V_(2)V_{2}V2. Do not confuse this 2 -form with the differential 2form, ω ω omega\omegaω, of Equation 5.2. The 2 -form v is only defined at the single tangent space T p R 3 T p R 3 T_(p)R^(3)T_{p} \mathrm{R}^{3}TpR3, whereas ω ω omega\omegaω is defined everywhere.
Let us now see how we can use v to decide whether to use the " + " or "-" sign in Equation 5.2. All we must do is calculate ν ( ϕ x ( x p , y p ) , ϕ y ( x p , y p ) ) ν ϕ x x p , y p , ϕ y x p , y p nu((del phi)/(del x)(x_(p),y_(p)),(del phi)/(del y)(x_(p),y_(p)))\nu\left(\frac{\partial \phi}{\partial x}\left(x_{p}, y_{p}\right), \frac{\partial \phi}{\partial y}\left(x_{p}, y_{p}\right)\right)ν(ϕx(xp,yp),ϕy(xp,yp)), where φ ( x p , y p ) = p φ x p , y p = p varphi(x_(p),y_(p))=p\varphi\left(x_{p}, y_{p}\right)=pφ(xp,yp)=p. If the result is positive, then we will use the " + " sign to alculate the integral in Equation 5.2. If it is negative then we use the "-" sign. Let's see how this works with an example.
Example 21. Let's revisit Example 20. The problem was to integrate the form z 2 d x d y z 2 d x d y z^(2)dx^^dyz^{2} d x \wedge d yz2dxdy over M M MMM, the top half of the unit sphere. But no orientation was ever given for M M MMM, so the problem was not very well stated. Let's pick an easy point, p p ppp, on M : ( 0 , 2 / 2 , 2 / 2 ) M : ( 0 , 2 / 2 , 2 / 2 ) M:(0,sqrt()2//2,sqrt()2//2)M:(0, \sqrt{ } 2 / 2, \sqrt{ } 2 / 2)M:(0,2/2,2/2). The vectors 1 , 0 , 0 p 1 , 0 , 0 p (:1,0,0:)_(p)\langle 1,0,0\rangle_{p}1,0,0p and 0 , 1 , 1 p 0 , 1 , 1 p (:0,1,-1:)_(p)\langle 0,1,-1\rangle_{p}0,1,1p in T p R 3 T p R 3 T_(p)R^(3)T_{p} \mathrm{R}^{3}TpR3 are both tangent to M M MMM. To give an orientation on M M MMM, all we do is specify a 2 -form vv\vee on T p R 3 T p R 3 T_(p)R^(3)T_{p} \mathrm{R}^{3}TpR3 such that V ( 1 , 0 , 0 , 0 , 1 , 1 ) 0 V ( 1 , 0 , 0 , 0 , 1 , 1 ) 0 V((:1,0,0:),(:0,1,-1:))!=0\mathrm{V}(\langle 1,0,0\rangle,\langle 0,1,-1\rangle) \neq 0V(1,0,0,0,1,1)0. Let's pick an easy one: v = d x v = d x v=dx\mathrm{v}=d xv=dx d y d y ^^dy\wedge d ydy.
Now, let's see what happens when we try to evaluate the integral by using the parameterization φ ( r t ) = ( r cos t , r sin t , 1 r 2 ) φ ( r t ) = r cos t , r sin t , 1 r 2 varphi^(')(rt)=(-r cos t,r sin t,sqrt()1-r^(2))\varphi^{\prime}(r t)=\left(-r \cos t, r \sin t, \sqrt{ } 1-r^{2}\right)φ(rt)=(rcost,rsint,1r2). First, note that φ ( 2 / 2 , π / 2 ) = ( 0 , 2 / 2 , 2 / 2 ) φ ( 2 / 2 , π / 2 ) = ( 0 , 2 / 2 , 2 / 2 ) varphi^(')(sqrt()2//2,pi//2)=(0,sqrt()2//2,sqrt()2//2)\varphi^{\prime}(\sqrt{ } 2 / 2, \pi / 2)=(0, \sqrt{ } 2 / 2, \sqrt{ } 2 / 2)φ(2/2,π/2)=(0,2/2,2/2) and
( ϕ r r 2 2 , π 2 ) , ϕ t ( 2 2 , π 2 ) ) = ( ( 0 , 1 , 1 , 2 2 , 0 , 0 ) . ϕ r r 2 2 , π 2 , ϕ t 2 2 , π 2 = ( 0 , 1 , 1 , 2 2 , 0 , 0 . {:((delphi^('))/(del r)r(sqrt2)/(2),(pi)/(2)),(delphi^('))/(del t)((sqrt2)/(2),(pi)/(2)))=((0,1,-1:),(:(sqrt2)/(2),0,0:)).\left.\left(\frac{\partial \phi^{\prime}}{\partial r} r \frac{\sqrt{2}}{2}, \frac{\pi}{2}\right), \frac{\partial \phi^{\prime}}{\partial t}\left(\frac{\sqrt{2}}{2}, \frac{\pi}{2}\right)\right)=\left((0,1,-1\rangle,\left\langle\frac{\sqrt{2}}{2}, 0,0\right\rangle\right) .(ϕrr22,π2),ϕt(22,π2))=((0,1,1,22,0,0).
Now we check the value of v when this pair is plugged in:
d x d y ( 0 , 1 , 1 , 2 2 , 0 , 0 ) = | 0 2 2 1 0 | = 2 2 . d x d y 0 , 1 , 1 , 2 2 , 0 , 0 = 0 2 2 1 0 = 2 2 . dx^^dy((:0,1,-1:),(:(sqrt2)/(2),0,0:))=|[0,(sqrt2)/(2)],[1,0]|=-(sqrt2)/(2).d x \wedge d y\left(\langle 0,1,-1\rangle,\left\langle\frac{\sqrt{2}}{2}, 0,0\right\rangle\right)=\left|\begin{array}{cc} 0 & \frac{\sqrt{2}}{2} \\ 1 & 0 \end{array}\right|=-\frac{\sqrt{2}}{2} .dxdy(0,1,1,22,0,0)=|02210|=22.
The sign of this result is "-," so we need to use the negative sign in Equation 5.2 in order to use φ φ varphi^(')\varphi^{\prime}φ to evaluate the integral of ω ω omega\omegaω over M M MMM.
M ω = R ω ϕ ( r , t ) ( ϕ r ( r , t ) , ϕ t ( r , t ) ) d r d t = R ( 1 r 2 ) | cos t r sin t sin t r cos t | d r d t = π 2 . M ω = R ω ϕ ( r , t ) ϕ r ( r , t ) , ϕ t ( r , t ) d r d t = R 1 r 2 cos t r sin t sin t r cos t d r d t = π 2 . {:[int_(M)omega=-int_(R)omega_(phi(r,t))((delphi^('))/(del r)(r,t),(delphi^('))/(del t)(r,t))dr^^dt],[=-int_(R)(1-r^(2))|[cos t,r sin t],[sin t,r cos t]|drdt=(pi)/(2).]:}\begin{aligned} \int_{M} \omega & =-\int_{R} \omega_{\phi(r, t)}\left(\frac{\partial \phi^{\prime}}{\partial r}(r, t), \frac{\partial \phi^{\prime}}{\partial t}(r, t)\right) d r \wedge d t \\ & =-\int_{R}\left(1-r^{2}\right)\left|\begin{array}{rr} \cos t & r \sin t \\ \sin t & r \cos t \end{array}\right| d r d t=\frac{\pi}{2} . \end{aligned}Mω=Rωϕ(r,t)(ϕr(r,t),ϕt(r,t))drdt=R(1r2)|costrsintsintrcost|drdt=π2.
Very often, the surface that we are going to integrate over is given to us by a parameterization. In this case, there is a very natural choice of orientation. Just use the "+' sign in Equation 5.2! We will call this the orientation of M M MMM induced by the parameterization. In other words, if you see a problem phrased like this, "Calculate the integral of the form ω ω omega\omegaω over the surface M M MMM given by parameterization φ φ varphi\varphiφ with the induced orientation," then you should just go back to using Equation 5.1 and do not worry about anything else. In fact, this situation is so common that when you are asked to integrate some form over a surface which is given by a parameterization, but no orientation is specified, then you should assume the induced orientation is the desired one.
5.9. Let M M MMM be the image of the parameterization, φ ( a , b ) = ( a , a + φ ( a , b ) = ( a , a + varphi(a,b)=(a,a+\varphi(a, b)=(a, a+φ(a,b)=(a,a+ b , a b ) b , a b ) b,ab)b, a b)b,ab), where 0 a 1 0 a 1 0 <= a <= 10 \leq a \leq 10a1 and 0 b 1 0 b 1 0 <= b <= 10 \leq b \leq 10b1. Integrate the form ω = 2 z ω = 2 z omega=2z\omega=2 zω=2z d x d y + y d y d z x d x d z d x d y + y d y d z x d x d z dx^^dy+ydy^^dz-xdx^^dzd x \wedge d y+y d y \wedge d z-x d x \wedge d zdxdy+ydydzxdxdz over M M MMM using the orientation induced by φ φ varphi\varphiφ.
There is one subtle technical point here that should be addressed. The novice reader may want to skip this for now. Suppose someone gives you a surface defined by a parameterization and tells you to integrate some 2 -form over it, using the induced orientation. But you are clever, and you realize that if you change parameterizations you
can make the integral easier. Which orientation do you use? The problem is that the orientation induced by your new parameterization may not be the same as the one induced by the original parameterization.
To fix this we need to see how we can define a 2 -form on some tangent space T p R 3 T p R 3 T_(p)R^(3)T_{p} \mathrm{R}^{3}TpR3, where p p ppp is a point of M , that yields an M , that yields an  M_(", that yields an ")M_{\text {, that yields an }}M, that yields an  orientation of M M MMM consistent with the one induced by a parameterization φ φ varphi\varphiφ. This is not so hard. If d x d y d x d y dx^^dyd x \wedge d ydxdy ( ϕ x ( x p , y p ) , ϕ y ( x p , y p ) ) is ϕ x x p , y p , ϕ y x p , y p is  ((del phi)/(del x)(x_(p),y_(p)),(del phi)/(del y)(x_(p),y_(p)))_("is ")\left(\frac{\partial \phi}{\partial x}\left(x_{p}, y_{p}\right), \frac{\partial \phi}{\partial y}\left(x_{p}, y_{p}\right)\right)_{\text {is }}(ϕx(xp,yp),ϕy(xp,yp))is  is positive, then we simply let v = d x d y v = d x d y v=dx^^dy\mathrm{v}=d x \wedge d yv=dxdy. If it is egative, then we let V = d x d y V = d x d y V=-dx^^dy\mathrm{V}=-d x \wedge d yV=dxdy. In the unlikely event that d x d y ( ϕ x ( x p , y p ) , ϕ y ( x p , y p ) ) = 0 d x d y ϕ x x p , y p , ϕ y x p , y p = 0 dx^^dy((del phi)/(del x)(x_(p),y_(p)),(del phi)/(del y)(x_(p),y_(p)))=0d x \wedge d y\left(\frac{\partial \phi}{\partial x}\left(x_{p}, y_{p}\right), \frac{\partial \phi}{\partial y}\left(x_{p}, y_{p}\right)\right)=0dxdy(ϕx(xp,yp),ϕy(xp,yp))=0 we can remedy things by either changing the point p p ppp or by using d x d z d x d z dx^^dzd x \wedge d zdxdz instead of d x d y d x d y dx^^dyd x \wedge d ydxdy. Once we have defined v v vvv, we know how to integrate M M MMM using any other parameterization.
5.10. Let Ψ Ψ Psi\PsiΨ be the following parameterization of the sphere of radius one:
ψ ( θ , ϕ ) = ( sin ϕ cos θ , sin ϕ sin θ , cos ϕ ) ψ ( θ , ϕ ) = ( sin ϕ cos θ , sin ϕ sin θ , cos ϕ ) psi(theta,phi)=(sin phi cos theta,sin phi sin theta,cos phi)\psi(\theta, \phi)=(\sin \phi \cos \theta, \sin \phi \sin \theta, \cos \phi)ψ(θ,ϕ)=(sinϕcosθ,sinϕsinθ,cosϕ)
Which of the following 2-forms on T ( 2 2 , 0 , 2 2 ) R 3 T 2 2 , 0 , 2 2 R 3 T_(((sqrt2)/(2),0,(sqrt2)/(2)))R^(3)T_{\left(\frac{\sqrt{2}}{2}, 0, \frac{\sqrt{2}}{2}\right)} \mathbb{R}^{3}T(22,0,22)R3 determine the same orientation on the sphere as that induced by Ψ Ψ Psi\PsiΨ ?
1. α = d x d y + 2 d y d z . 2. β = d x d y 2 d y d z . 3. γ = d x d z .  1.  α = d x d y + 2 d y d z  2.  β = d x d y 2 d y d z  3.  γ = d x d z . {:[" 1. "alpha=dx^^dy+2dy^^dz". "],[" 2. "beta=dx^^dy-2dy^^dz". "],[" 3. "gamma=dx^^dz.]:}\begin{aligned} & \text { 1. } \alpha=d x \wedge d y+2 d y \wedge d z \text {. } \\ & \text { 2. } \beta=d x \wedge d y-2 d y \wedge d z \text {. } \\ & \text { 3. } \gamma=d x \wedge d z . \end{aligned} 1. α=dxdy+2dydz 2. β=dxdy2dydz 3. γ=dxdz.

5.4 Integrating 1-forms on R m R m R^(m)\mathrm{R}^{\boldsymbol{m}}Rm

In the previous sections we saw how to integrate a 2 -form over a region in R 2 R 2 R^(2)R^{2}R2, or over a subset of R 3 R 3 R^(3)R^{3}R3 parameterized by a region in R 2 R 2 R^(2)R^{2}R2. The reason that these dimensions were chosen was because there is nothing new that needs to be introduced to move to the general case. In fact, if the reader were to go back and look at what we did, he/she would find that almost nothing would change if we had been talking about n n nnn-forms instead.
Before we jump to the general case, we will work one example showing how to integrate a 1 -form over a parameterized curve. Example 22. Let C C CCC be the curve in R 2 R 2 R^(2)\mathrm{R}^{2}R2 parameterized by
ϕ ( t ) = ( t 2 , t 3 ) ϕ ( t ) = t 2 , t 3 phi(t)=(t^(2),t^(3))\phi(t)=\left(t^{2}, t^{3}\right)ϕ(t)=(t2,t3)
where 0 t 2 0 t 2 0 <= t <= 20 \leq t \leq 20t2. Let V be the 1 -form y d x + x d y y d x + x d y ydx+xdyy d x+x d yydx+xdy. We calculate C V C V int_(C)V\int_{\mathrm{C}} \mathrm{V}CV.
The first step is to calculate
d ϕ d t = 2 t , 3 t 2 . d ϕ d t = 2 t , 3 t 2 . (d phi)/(dt)=(:2t,3t^(2):).\frac{d \phi}{d t}=\left\langle 2 t, 3 t^{2}\right\rangle .dϕdt=2t,3t2.
So, d x = 2 t d x = 2 t dx=2td x=2 tdx=2t and d y = 3 t 2 d y = 3 t 2 dy=3t^(2)d y=3 t^{2}dy=3t2. From the parameterization we also know x = t 2 x = t 2 x=t^(2)x=t^{2}x=t2 and y = t 3 y = t 3 y=t^(3)y=t^{3}y=t3. Hence, since v = y d x + x d y v = y d x + x d y v=ydx+xdy\mathrm{v}=y d x+x d yv=ydx+xdy, we have
v ϕ ( t ) ( d ϕ d t ) = ( t 3 ) ( 2 t ) + ( t 2 ) ( 3 t 2 ) = 5 t 4 . v ϕ ( t ) d ϕ d t = t 3 ( 2 t ) + t 2 3 t 2 = 5 t 4 . v_(phi(t))((d phi)/(dt))=(t^(3))(2t)+(t^(2))(3t^(2))=5t^(4).v_{\phi(t)}\left(\frac{d \phi}{d t}\right)=\left(t^{3}\right)(2 t)+\left(t^{2}\right)\left(3 t^{2}\right)=5 t^{4} .vϕ(t)(dϕdt)=(t3)(2t)+(t2)(3t2)=5t4.
Finally, we integrate:
C v = 0 2 v ϕ ( t ) ( d ϕ d t ) d t = 0 2 5 t 4 d t = t 5 | 0 2 = 32 C v = 0 2 v ϕ ( t ) d ϕ d t d t = 0 2 5 t 4 d t = t 5 0 2 = 32 {:[int_(C)v=int_(0)^(2)v_(phi(t))((d phi)/(dt))dt],[=int_(0)^(2)5t^(4)dt],[=t^(5)|_(0)^(2)],[=32]:}\begin{aligned} \int_{C} v & =\int_{0}^{2} v_{\phi(t)}\left(\frac{d \phi}{d t}\right) d t \\ & =\int_{0}^{2} 5 t^{4} d t \\ & =\left.t^{5}\right|_{0} ^{2} \\ & =32 \end{aligned}Cv=02vϕ(t)(dϕdt)dt=025t4dt=t5|02=32
5.11. Let C C CCC be the curve in R 3 R 3 R^(3)\mathrm{R}^{3}R3 parameterized by φ ( t ) = ( t , t 2 , 1 + φ ( t ) = t , t 2 , 1 + varphi(t)=(t,t^(2),1+:}\varphi(t)=\left(t, t^{2}, 1+\right.φ(t)=(t,t2,1+ t t ttt ), where 0 t 2 0 t 2 0 <= t <= 20 \leq t \leq 20t2. Integrate the 1-form ω = y d x + z d y + x y d z ω = y d x + z d y + x y d z omega=ydx+zdy+xydz\omega=y d x+z d y+x y d zω=ydx+zdy+xydz over C C CCC using the induced orientation.
5.12. Let C C CCC be the curve parameterized by the following:
ϕ ( t ) = ( 2 cos t , 2 sin t , t 2 ) , 0 t 2 . ϕ ( t ) = 2 cos t , 2 sin t , t 2 , 0 t 2 . phi(t)=(2cos t,2sin t,t^(2)),quad0 <= t <= 2.\phi(t)=\left(2 \cos t, 2 \sin t, t^{2}\right), \quad 0 \leq t \leq 2 .ϕ(t)=(2cost,2sint,t2),0t2.
Integrate the 1-form ( x 2 + y 2 ) d z x 2 + y 2 d z (x^(2)+y^(2))dz\left(x^{2}+y^{2}\right) d z(x2+y2)dz over C C CCC.
5.13. Let C C CCC be the subset of the graph of y = x 2 y = x 2 y=x^(2)y=x^{2}y=x2 where 0 x 1 0 x 1 0 <= x <= 10 \leq x \leq 10x1. An orientation on C C CCC is given by the 1-form d x d x dxd xdx on T ( 0 , 0 ) R 2 T ( 0 , 0 ) R 2 T_((0,0))R^(2)T_{(0,0)} \mathrm{R}^{2}T(0,0)R2. Let ω ω omega\omegaω be the 1 -form x 4 d x + x y d y x 4 d x + x y d y -x^(4)dx+xydy-x^{4} d x+x y d yx4dx+xydy. Integrate ω ω omega\omegaω over C C CCC.
5.14. Let M M MMM be the line segment in R 2 R 2 R^(2)R^{2}R2 which connects ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0) to (4, 6). An orientation on M M MMM is specified by the 1 -form d x d x -dx-d xdx on T ( 2 , 3 ) R 2 T ( 2 , 3 ) R 2 T_((2,3))R^(2)T_{(2,3)} \mathrm{R}^{2}T(2,3)R2. Integrate the form ω = sin y d x + cos x d y ω = sin y d x + cos x d y omega=sin ydx+cos xdy\omega=\sin y d x+\cos x d yω=sinydx+cosxdy over M M MMM.
Just as there was for surfaces, for parameterized curves there is also a pictorial way to specify an orientation. All we have to do is place an arrowhead somewhere along the curve, and ask whether or
not the derivative of the parameterization gives a tangent vector that points in the same direction. We illustrate this in the next example.
Example 23. Let C C CCC be the portion of the graph of x = y 2 x = y 2 x=y^(2)x=y^{2}x=y2 where 0 x 0 x 0 <= x0 \leq x0x 1 1 <= 1\leq 11, as pictured in Figure 5.6. Notice the arrowhead on C C CCC. We integrate the 1 -form ω = d x + d y ω = d x + d y omega=dx+dy\omega=d x+d yω=dx+dy over C C CCC with the indicated orientation.
First, parameterize C C CCC as φ ( t ) = ( t 2 , t ) φ ( t ) = t 2 , t varphi(t)=(t^(2),t)\varphi(t)=\left(t^{2}, t\right)φ(t)=(t2,t), where 0 t 1 0 t 1 0 <= t <= 10 \leq t \leq 10t1. Now notice that the derivative of φ φ varphi\varphiφ is
Fig. 5.6. An orientation on C C CCC is given by an arrowhead.
d ϕ d t = 2 t , 1 . d ϕ d t = 2 t , 1 . (d phi)/(dt)=(:2t,1:).\frac{d \phi}{d t}=\langle 2 t, 1\rangle .dϕdt=2t,1.
At the point ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0) this is the vector 0 , 1 0 , 1 (:0,1:)\langle 0,1\rangle0,1, which points in a direction opposite to that of the arrowhead. This tells us to use a negative sign when we integrate, as follows:
C ω = 0 1 ω ( t 2 , t ) ( 2 t , 1 ) = ( 2 t + 1 ) | 0 1 = 2 . C ω = 0 1 ω t 2 , t ( 2 t , 1 ) = ( 2 t + 1 ) 0 1 = 2 . {:[int_(C)omega=-int_(0)^(1)omega_((t^(2),t))((:2t","1:))],[=-(2t+1)|_(0)^(1)],[=-2.]:}\begin{aligned} \int_{C} \omega & =-\int_{0}^{1} \omega_{\left(t^{2}, t\right)}(\langle 2 t, 1\rangle) \\ & =-\left.(2 t+1)\right|_{0} ^{1} \\ & =-2 . \end{aligned}Cω=01ω(t2,t)(2t,1)=(2t+1)|01=2.

5.5 Integrating n n n\boldsymbol{n}n-forms on R m R m R^(m)\mathrm{R}^{\boldsymbol{m}}Rm

To proceed to the general case, we need to know what the integral of an n n nnn-form over a region of R n R n R^(n)\mathrm{R}^{n}Rn is. The steps to define such an object are precisely the same as before, and the results are similar. If our coordinates in R n R n R^(n)\mathrm{R}^{n}Rn are ( x 1 , x 2 , , x n x 1 , x 2 , , x n x_(1),x_(2),dots,x_(n)x_{1}, x_{2}, \ldots, x_{n}x1,x2,,xn ), then an n n nnn-form is always given by
f ( x 1 , , x n ) d x 1 d x 2 d x n f x 1 , , x n d x 1 d x 2 d x n f(x_(1),dots,x_(n))dx_(1)^^dx_(2)^^dots^^dx_(n)f\left(x_{1}, \ldots, x_{n}\right) d x_{1} \wedge d x_{2} \wedge \ldots \wedge d x_{n}f(x1,,xn)dx1dx2dxn
Going through the steps, we find that the definition of ω ω int omega\int \omegaω is exactly the same as the definition we learned in Chapter 1 for R n f d x 1 d x 2 d x n R n f d x 1 d x 2 d x n int_(R^(n))fdx_(1)dx_(2)dots dx_(n)\int_{\mathbb{R}^{n}} f d x_{1} d x_{2} \ldots d x_{n}Rnfdx1dx2dxn
5.15. Let Ω Ω Omega\OmegaΩ be the cube in R 3 R 3 R^(3)\mathrm{R}^{3}R3
{ ( x , y , z ) 0 x , y , z 1 } . { ( x , y , z ) 0 x , y , z 1 } . {(x,y,z)∣0 <= x,y,z <= 1}.\{(x, y, z) \mid 0 \leq x, y, z \leq 1\} .{(x,y,z)0x,y,z1}.
Let y y yyy be the 3-form x 2 z d x d y d z x 2 z d x d y d z x^(2)zdx^^dy^^dzx^{2} z d x \wedge d y \wedge d zx2zdxdydz. Calculate γ . Ω γ . Ω int gamma.Omega\int \gamma . \Omegaγ.Ω
Moving on to integrals of n n nnn-forms over subsets of R m R m R^(m)\mathrm{R}^{m}Rm parameterized by a region in R n R n R^(n)\mathrm{R}^{n}Rn, we again find nothing surprising. Suppose we denote such a subset by M M MMM. Let φ : R R n M R m φ : R R n M R m varphi:R subR^(n)rarr M subR^(m)\varphi: R \subset \mathrm{R}^{n} \rightarrow M \subset \mathrm{R}^{m}φ:RRnMRm
be a parameterization. Then we find that the following generalization of Equation 5.2 must hold:
(5.3) M ω = ± R ω ϕ ( x 1 , , x n ) ( ϕ x 1 ( x 1 , x n ) , , ϕ x n ( x 1 , x n ) ) d x 1 d x n (5.3) M ω = ± R ω ϕ x 1 , , x n ϕ x 1 x 1 , x n , , ϕ x n x 1 , x n d x 1 d x n {:(5.3)int_(M)omega=+-int_(R)omega_(phi(x_(1),dots,x_(n)))((del phi)/(delx_(1))(x_(1),dotsx_(n)),dots,(del phi)/(delx_(n))(x_(1),dotsx_(n)))dx_(1)^^dots^^dx_(n):}\begin{equation*} \int_{M} \omega= \pm \int_{R} \omega_{\phi\left(x_{1}, \ldots, x_{n}\right)}\left(\frac{\partial \phi}{\partial x_{1}}\left(x_{1}, \ldots x_{n}\right), \ldots, \frac{\partial \phi}{\partial x_{n}}\left(x_{1}, \ldots x_{n}\right)\right) d x_{1} \wedge \ldots \wedge d x_{n} \tag{5.3} \end{equation*}(5.3)Mω=±Rωϕ(x1,,xn)(ϕx1(x1,xn),,ϕxn(x1,xn))dx1dxn
To decide whether or not to use the negative sign in this equation we must specify an orientation. Again, one way to do this is to give an n n nnn-form V on T p R m T p R m T_(p)R^(m)T_{p} \mathrm{R}^{m}TpRm, where p p ppp is some point of M M MMM. We use the negative sign when the value of is negative, where φ ( x 1 , x n ) = p φ x 1 , x n = p varphi(x_(1),dotsx_(n))=p\varphi\left(x_{1}, \ldots x_{n}\right)=pφ(x1,xn)=p. If M M MMM was originally given by a parameterization and we are instructed to use the induced orientation, then we can ignore the negative sign.
v ( ϕ x 1 ( x 1 , x n ) , , ϕ x n ( x 1 , x n ) ) v ϕ x 1 x 1 , x n , , ϕ x n x 1 , x n v((del phi)/(delx_(1))(x_(1),dotsx_(n)),dots,(del phi)/(delx_(n))(x_(1),dotsx_(n)))v\left(\frac{\partial \phi}{\partial x_{1}}\left(x_{1}, \ldots x_{n}\right), \ldots, \frac{\partial \phi}{\partial x_{n}}\left(x_{1}, \ldots x_{n}\right)\right)v(ϕx1(x1,xn),,ϕxn(x1,xn))
Example 24. Suppose φ ( a , b , c ) = ( a + b , a + c , b c , a 2 ) φ ( a , b , c ) = a + b , a + c , b c , a 2 varphi(a,b,c)=(a+b,a+c,bc,a^(2))\varphi(a, b, c)=\left(a+b, a+c, b c, a^{2}\right)φ(a,b,c)=(a+b,a+c,bc,a2), where 0 0 0 <=0 \leq0 a , b , c 1 a , b , c 1 a,b,c <= 1a, b, c \leq 1a,b,c1. Let M M MMM be the image of φ φ varphi\varphiφ with the induced orientation. Suppose ω = d y d z d w d x d z d w 2 y d x d y d z ω = d y d z d w d x d z d w 2 y d x d y d z omega=dy^^dz^^dw-dx^^dz^^dw-2ydx^^dy^^dz\omega=d y \wedge d z \wedge d w-d x \wedge d z \wedge d w-2 y d x \wedge d y \wedge d zω=dydzdwdxdzdw2ydxdydz. Then,
M ω = R ω ϕ ( a , b , c ) ( ϕ a ( a , b , c ) , ϕ b ( a , b , c ) , ϕ c ( a , b , c ) ) d a d b d c = R ω ϕ ( a , b , c ) ( 1 , 1 , 0 , 2 a , 1 , 0 , c , 0 , 0 , 1 , b , 0 ) d a d b d c = R | 1 0 1 0 c b 2 a 0 0 | | 1 1 0 0 c b 2 a 0 0 | 2 ( a + c ) | 1 1 0 1 0 1 0 c b | d a d b d c = 0 1 0 1 0 1 2 b c + 2 c 2 d a d b d c = 7 6 . M ω = R ω ϕ ( a , b , c ) ϕ a ( a , b , c ) , ϕ b ( a , b , c ) , ϕ c ( a , b , c ) d a d b d c = R ω ϕ ( a , b , c ) ( 1 , 1 , 0 , 2 a , 1 , 0 , c , 0 , 0 , 1 , b , 0 ) d a d b d c = R 1 0 1 0 c b 2 a 0 0 1 1 0 0 c b 2 a 0 0 2 ( a + c ) 1 1 0 1 0 1 0 c b d a d b d c = 0 1 0 1 0 1 2 b c + 2 c 2 d a d b d c = 7 6 . {:[int_(M)omega=int_(R)omega_(phi(a,b,c))((del phi)/(del a)(a,b,c),(del phi)/(del b)(a,b,c),(del phi)/(del c)(a,b,c))da^^db^^dc],[=int_(R)omega_(phi(a,b,c))((:1","1","0","2a:)","(:1","0","c","0:)","(:0","1","b","0:))da^^db^^dc],[=int_(R)|[1,0,1],[0,c,b],[2a,0,0]|-|[1,1,0],[0,c,b],[2a,0,0]|-2(a+c)|[1,1,0],[1,0,1],[0,c,b]|da^^db^^dc],[=int_(0)^(1)int_(0)^(1)int_(0)^(1)2bc+2c^(2)dadbdc=(7)/(6).]:}\begin{aligned} \int_{M} \omega & =\int_{R} \omega_{\phi(a, b, c)}\left(\frac{\partial \phi}{\partial a}(a, b, c), \frac{\partial \phi}{\partial b}(a, b, c), \frac{\partial \phi}{\partial c}(a, b, c)\right) d a \wedge d b \wedge d c \\ & =\int_{R} \omega_{\phi(a, b, c)}(\langle 1,1,0,2 a\rangle,\langle 1,0, c, 0\rangle,\langle 0,1, b, 0\rangle) d a \wedge d b \wedge d c \\ & =\int_{R}\left|\begin{array}{rrr} 1 & 0 & 1 \\ 0 & c & b \\ 2 a & 0 & 0 \end{array}\right|-\left|\begin{array}{rrr} 1 & 1 & 0 \\ 0 & c & b \\ 2 a & 0 & 0 \end{array}\right|-2(a+c)\left|\begin{array}{lll} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & c & b \end{array}\right| d a \wedge d b \wedge d c \\ & =\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} 2 b c+2 c^{2} d a d b d c=\frac{7}{6} . \end{aligned}Mω=Rωϕ(a,b,c)(ϕa(a,b,c),ϕb(a,b,c),ϕc(a,b,c))dadbdc=Rωϕ(a,b,c)(1,1,0,2a,1,0,c,0,0,1,b,0)dadbdc=R|1010cb2a00||1100cb2a00|2(a+c)|1101010cb|dadbdc=0101012bc+2c2dadbdc=76.

5.6 The change of variables formula

There is a special case of Equation 5.3 which is worth noting. Suppose Ψ Ψ Psi\PsiΨ is a parameterization that takes some subregion, R R RRR, of R n R n R^(n)\mathrm{R}^{n}Rn into some other subregion, M M MMM, of R n R n R^(n)\mathrm{R}^{n}Rn and ω ω omega\omegaω is an n n nnn-form on R n R n R^(n)\mathrm{R}^{n}Rn. At each point, ω ω omega\omegaω is just a volume form, so it can be written as f ( x 1 , f x 1 , f(x_(1),dots:}f\left(x_{1}, \ldots\right.f(x1,, x n ) d x 1 d x n x n d x 1 d x n {:x_(n))dx_(1)^^dots^^dx_(n)\left.x_{n}\right) d x_{1} \wedge \ldots \wedge d x_{n}xn)dx1dxn. If we let x = ( x 1 , x n ) x = x 1 , x n x=(x_(1),dotsx_(n))\mathrm{x}=\left(x_{1}, \ldots x_{n}\right)x=(x1,xn) then Equation 5.3 can be written as:
(5.4) M f ( x ¯ ) d x 1 d x n = ± R f ( ϕ ( x ¯ ) ) | ϕ x 1 ( x ¯ ) ϕ x n ( x ¯ ) | d x 1 d x n (5.4) M f ( x ¯ ) d x 1 d x n = ± R f ( ϕ ( x ¯ ) ) ϕ x 1 ( x ¯ ) ϕ x n ( x ¯ ) d x 1 d x n {:(5.4)int_(M)f( bar(x))dx_(1)dots dx_(n)=+-int_(R)f(phi( bar(x)))|(del phi)/(delx_(1))(( bar(x)))dots(del phi)/(delx_(n))(( bar(x)))|dx_(1)dots dx_(n):}\begin{equation*} \int_{M} f(\bar{x}) d x_{1} \ldots d x_{n}= \pm \int_{R} f(\phi(\bar{x}))\left|\frac{\partial \phi}{\partial x_{1}}(\bar{x}) \ldots \frac{\partial \phi}{\partial x_{n}}(\bar{x})\right| d x_{1} \ldots d x_{n} \tag{5.4} \end{equation*}(5.4)Mf(x¯)dx1dxn=±Rf(ϕ(x¯))|ϕx1(x¯)ϕxn(x¯)|dx1dxn
The bars | | | | |*||\cdot||| indicate that we take the determinant of the matrix whose column vectors are ϕ x i ( x ¯ ) ϕ x i ( x ¯ ) (del phi)/(delx_(i))( bar(x))\frac{\partial \phi}{\partial x_{i}}(\bar{x})ϕxi(x¯).

5.6.1 1-forms on 1 1 ^(1){ }^{1}1

When n = 1 n = 1 n=1n=1n=1 this is just the substitution rule for integration from calculus. We demonstrate this in the following example.
Example 25. Let's integrate the 1 -form ω = u d u ω = u d u omega=sqrt()udu\omega=\sqrt{ } u d uω=udu over the interval [ 1 , 5 ] [ 1 , 5 ] [1,5][1,5][1,5]. This would be easy enough to do directly, but using a parameterization of this interval will be instructive. Let φ : [ 0 , 2 ] φ : [ 0 , 2 ] varphi:[0,2]rarr\varphi:[0,2] \rightarrowφ:[0,2] [ 1 , 5 ] [ 1 , 5 ] [1,5][1,5][1,5] be the parameterization given by φ ( x ) = x 2 + 1 φ ( x ) = x 2 + 1 varphi(x)=x^(2)+1\varphi(x)=x^{2}+1φ(x)=x2+1. Then d ϕ d x = 2 x d ϕ d x = 2 x (d phi)/(dx)=2x\frac{d \phi}{d x}=2 xdϕdx=2x . Now we compute:
1 5 u d u = [ 1 , 5 ] ω = [ 0 , 2 ] ω ϕ ( x ) ( d ϕ d x ) d x = [ 0 , 2 ] ω x 2 + 1 ( ( 2 x ) d x = [ 0 , 2 ] 2 2 x x 2 + 1 d x = 0 2 2 x x 2 + 1 d x 1 5 u d u = [ 1 , 5 ] ω = [ 0 , 2 ] ω ϕ ( x ) d ϕ d x d x = [ 0 , 2 ] ω x 2 + 1 ( ( 2 x ) d x = [ 0 , 2 ] 2 2 x x 2 + 1 d x = 0 2 2 x x 2 + 1 d x {:[int_(1)^(5)sqrtudu=int_([1,5])omega=int_([0,2])omega_(phi(x))((d phi)/(dx))dx],[=int_([0,2])omega_(x^(2)+1)((2x:))dx],[=int_([0,2])^(2)2xsqrt(x^(2)+1)dx],[=int_(0)^(2)2xsqrt(x^(2)+1)dx]:}\begin{aligned} \int_{1}^{5} \sqrt{u} d u=\int_{[1,5]} \omega & =\int_{[0,2]} \omega_{\phi(x)}\left(\frac{d \phi}{d x}\right) d x \\ & =\int_{[0,2]} \omega_{x^{2}+1}((2 x\rangle) d x \\ & =\int_{[0,2]}^{2} 2 x \sqrt{x^{2}+1} d x \\ & =\int_{0}^{2} 2 x \sqrt{x^{2}+1} d x \end{aligned}15udu=[1,5]ω=[0,2]ωϕ(x)(dϕdx)dx=[0,2]ωx2+1((2x)dx=[0,2]22xx2+1dx=022xx2+1dx
Reading this backwards is doing the integral0 0 2 2 x x 2 + 1 d x 0 2 2 x x 2 + 1 d x int_(0)^(2)2xsqrt(x^(2)+1)dx\int_{0}^{2} 2 x \sqrt{x^{2}+1} d x022xx2+1dx by " u u uuu - - substitution" substitution."

5.6.2 2-forms on 2 2 ^(2){ }^{2}2

For other n n nnn, Equation 5.4 is the general change of variables formula.
Example 26. We will use the parameterization Ψ ( u , U ) = ( u , u 2 + U 2 ) Ψ ( u , U ) = u , u 2 + U 2 Psi(u,U)=(u,u^(2)+U^(2))\Psi(u, U)=\left(u, u^{2}+U^{2}\right)Ψ(u,U)=(u,u2+U2) to evaluate where R R RRR is the region of the x y x y xyx yxy-plane bounded by the parabolas y = x 2 y = x 2 y=x^(2)y=x^{2}y=x2 and y = x 2 + 4 y = x 2 + 4 y=x^(2)+4y=x^{2}+4y=x2+4, and the lines x = 0 x = 0 x=0x=0x=0 and x = 1 x = 1 x=1x=1x=1.
R ( x 2 + y ) d A R x 2 + y d A ∬_(R)(x^(2)+y)dA\iint_{R}\left(x^{2}+y\right) d AR(x2+y)dA
The first step is to find out what the limits of integration will be when we change coordinates.
y = x 2 u 2 + v 2 = u 2 v = 0 y = x 2 + 4 u 2 + v 2 = u 2 + 4 v = 2 x = 0 u = 0 x = 1 u = 1 . y = x 2 u 2 + v 2 = u 2 v = 0 y = x 2 + 4 u 2 + v 2 = u 2 + 4 v = 2 x = 0 u = 0 x = 1 u = 1 . {:[y=x^(2)=>u^(2)+v^(2)=u^(2)=>v=0],[y=x^(2)+4=>u^(2)+v^(2)=u^(2)+4=>v=2],[x=0=>u=0],[x=1=>u=1.]:}\begin{gathered} y=x^{2} \Rightarrow u^{2}+v^{2}=u^{2} \Rightarrow v=0 \\ y=x^{2}+4 \Rightarrow u^{2}+v^{2}=u^{2}+4 \Rightarrow v=2 \\ x=0 \Rightarrow u=0 \\ x=1 \Rightarrow u=1 . \end{gathered}y=x2u2+v2=u2v=0y=x2+4u2+v2=u2+4v=2x=0u=0x=1u=1.
Next, we will need the partial derivatives.
Ψ u = 1 , 2 u Ψ v = 0 , 2 v . Ψ u = 1 , 2 u Ψ v = 0 , 2 v . {:[(del Psi)/(del u)=(:1","2u:)],[(del Psi)/(del v)=(:0","2v:).]:}\begin{aligned} & \frac{\partial \Psi}{\partial u}=\langle 1,2 u\rangle \\ & \frac{\partial \Psi}{\partial v}=\langle 0,2 v\rangle . \end{aligned}Ψu=1,2uΨv=0,2v.
Finally, we can integrate.
R ( x 2 + y ) d A = R ( x 2 + y ) d x d y = 0 2 0 1 u 2 + ( u 2 + v 2 ) | 1 0 2 u 2 v | d u d v = 0 2 0 1 4 v u 2 + 2 v 3 d u d v = 0 2 4 3 v + 2 v 3 d v = 8 3 + 8 = 32 3 R x 2 + y d A = R x 2 + y d x d y = 0 2 0 1 u 2 + u 2 + v 2 1 0 2 u 2 v d u d v = 0 2 0 1 4 v u 2 + 2 v 3 d u d v = 0 2 4 3 v + 2 v 3 d v = 8 3 + 8 = 32 3 {:[∬_(R)(x^(2)+y)dA=int_(R)(x^(2)+y)dx^^dy],[=int_(0)^(2)int_(0)^(1)u^(2)+(u^(2)+v^(2))|[1,0],[2u,2v]|dudv],[=int_(0)^(2)int_(0)^(1)4vu^(2)+2v^(3)dudv],[=int_(0)^(2)(4)/(3)v+2v^(3)dv],[=(8)/(3)+8=(32)/(3)]:}\begin{aligned} \iint_{R}\left(x^{2}+y\right) d A & =\int_{R}\left(x^{2}+y\right) d x \wedge d y \\ & =\int_{0}^{2} \int_{0}^{1} u^{2}+\left(u^{2}+v^{2}\right)\left|\begin{array}{rr} 1 & 0 \\ 2 u & 2 v \end{array}\right| d u d v \\ & =\int_{0}^{2} \int_{0}^{1} 4 v u^{2}+2 v^{3} d u d v \\ & =\int_{0}^{2} \frac{4}{3} v+2 v^{3} d v \\ & =\frac{8}{3}+8=\frac{32}{3} \end{aligned}R(x2+y)dA=R(x2+y)dxdy=0201u2+(u2+v2)|102u2v|dudv=02014vu2+2v3dudv=0243v+2v3dv=83+8=323
Example 27. In our second example, we revisit Fubini's theorem, which says that the order of integration does not matter in a multiple integral. Recall from Section 5.2 the curious fact that f d x d y = f f d x d y = f int fdxdy=int f\int f d x d y=\int ffdxdy=f d x d y d x d y dx^^dyd x \wedge d ydxdy, but f d y d x f d y d x f d y d x f d y d x int fdydx!=int fdy^^dx\int f d y d x \neq \int f d y \wedge d xfdydxfdydx. We are now prepared to see why this s.
Let's suppose we want to integrate the function f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) over the rectangle R R RRR in R 2 R 2 R^(2)\mathrm{R}^{2}R2 with vertices at ( 0,0 ), ( a , 0 a , 0 a,0a, 0a,0 ), ( 0 , b 0 , b 0,b0, b0,b ) and ( a , b a , b a,ba, ba,b ). We know the answer is just
0 b 0 a f ( x , y ) d x d y 0 b 0 a f ( x , y ) d x d y int_(0)^(b)int_(0)^(a)f(x,y)dxdy\int_{0}^{b} \int_{0}^{a} f(x, y) d x d y0b0af(x,y)dxdy.
. We also know this integral is equal to R f d x R f d x int_(R)fdx^^\int_{\mathrm{R}} f d x \wedgeRfdx d y d y dyd ydy, where R R RRR s given the "standard" orientation (e.g., the one specifie by a counter-clockwise oriented circle).
Let's see what happens if we try to compute the integral using the following parameterization:
ϕ ( y , x ) = ( x , y ) , 0 y b , 0 x a . ϕ ( y , x ) = ( x , y ) , 0 y b , 0 x a . phi(y,x)=(x,y),0 <= y <= b,0 <= x <= a.\phi(y, x)=(x, y), 0 \leq y \leq b, 0 \leq x \leq a .ϕ(y,x)=(x,y),0yb,0xa.
First, we need the partials of φ φ varphi\varphiφ :
ϕ y = 0 , 1 ϕ x = 1 , 0 . ϕ y = 0 , 1 ϕ x = 1 , 0 . {:[(del phi)/(del y)=(:0","1:)],[(del phi)/(del x)=(:1","0:).]:}\begin{aligned} & \frac{\partial \phi}{\partial y}=\langle 0,1\rangle \\ & \frac{\partial \phi}{\partial x}=\langle 1,0\rangle . \end{aligned}ϕy=0,1ϕx=1,0.
Next we have to deal with the issue of orientation. The pair of vectors we just found, 0 , 1 0 , 1 (:0,1:)\langle 0,1\rangle0,1 and 1 , 0 1 , 0 (:1,0:)\langle 1,0\rangle1,0 are in an order which does not agree with the orientation of R R RRR. So we have to use the negative sign when employing Equation 5.4:
R f ( x , y ) d x d y = R f ( ϕ ( y , x ) ) | ϕ y ϕ x | d y d x = R f ( x , y ) | 0 1 1 0 | d y d x = R f ( x , y ) ( 1 ) d y d x = R f ( x , y ) d y d x . R f ( x , y ) d x d y = R f ( ϕ ( y , x ) ) ϕ y ϕ x d y d x = R f ( x , y ) 0 1 1 0 d y d x = R f ( x , y ) ( 1 ) d y d x = R f ( x , y ) d y d x . {:[int_(R)f(x","y)dxdy=-int_(R)f(phi(y","x))|(del phi)/(del y)(del phi)/(del x)|dydx],[=-int_(R)f(x","y)|[0,1],[1,0]|dy^^dx],[=-int_(R)f(x","y)(-1)dy^^dx],[=int_(R)f(x","y)dydx.]:}\begin{aligned} \int_{R} f(x, y) d x d y & =-\int_{R} f(\phi(y, x))\left|\frac{\partial \phi}{\partial y} \frac{\partial \phi}{\partial x}\right| d y d x \\ & =-\int_{R} f(x, y)\left|\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right| d y \wedge d x \\ & =-\int_{R} f(x, y)(-1) d y \wedge d x \\ & =\int_{R} f(x, y) d y d x . \end{aligned}Rf(x,y)dxdy=Rf(ϕ(y,x))|ϕyϕx|dydx=Rf(x,y)|0110|dydx=Rf(x,y)(1)dydx=Rf(x,y)dydx.
From the above, we see one of the reasons why Fubini's theorem is true is because when the order of integration is switched there are two negative signs. So, R f d y d x R f d y d x int_(R)fdydx\int_{\mathrm{R}} f d y d xRfdydx actually does equal R f d y d x R f d y d x int_(R)fdy^^dx\int_{R} f d y \wedge d xRfdydx but only if you remember to swit h h hhh the orientation of R R RRR !
5.16. Let E E EEE be the region in R 2 R 2 R^(2)R^{2}R2 parameterized by Ψ ( u , U ) = ( u 2 + U Ψ ( u , U ) = u 2 + U Psi(u,U)=(u^(2)+U:}\Psi(u, U)=\left(u^{2}+U\right.Ψ(u,U)=(u2+U 2 , 2 u ( u ) 2 , 2 u ( u ) ^(2),2u(u){ }^{2}, 2 u(u)2,2u(u), where 0 u 1 0 u 1 0 <= u <= 10 \leq u \leq 10u1 and 0 U 1 0 U 1 0 <= U <= 10 \leq U \leq 10U1. Evaluate
E 1 x y d A . E 1 x y d A . int_(E)(1)/(sqrt(x-y))dA.\int_{E} \frac{1}{\sqrt{x-y}} d A .E1xydA.
Up until this point, we have only seen how to integrate functions f f fff ( x , y ) ( x , y ) (x,y)(x, y)(x,y) over regions in the plane which are rectangles. Let's now see how we can use parameterizations to integrate over more general regions. Suppose first, that R R RRR is the region of the plane below the graph of y = g ( x ) y = g ( x ) y=g(x)y=g(x)y=g(x), above the x x xxx-axis, and between the lines x = a x = a x=ax=ax=a and x = b x = b x=bx=bx=b.
The region R R RRR can be parameterized by where a u b a u b a <= u <= ba \leq u \leq baub and 0 U 0 U 0 <= U0 \leq U0U 1 1 <= 1\leq 11. The partials of this parameterization are
Ψ ( u , v ) = ( u , v g ( u ) ) Ψ u = 1 , v d g ( u ) d u Ψ v = 0 , g ( u ) Ψ ( u , v ) = ( u , v g ( u ) ) Ψ u = 1 , v d g ( u ) d u Ψ v = 0 , g ( u ) {:[Psi(u","v)=(u","vg(u))],[(del Psi)/(del u)=(:1,v(dg(u))/(du):)],[(del Psi)/(del v)=(:0","g(u):)]:}\begin{gathered} \Psi(u, v)=(u, v g(u)) \\ \frac{\partial \Psi}{\partial u}=\left\langle 1, v \frac{d g(u)}{d u}\right\rangle \\ \frac{\partial \Psi}{\partial v}=\langle 0, g(u)\rangle \end{gathered}Ψ(u,v)=(u,vg(u))Ψu=1,vdg(u)duΨv=0,g(u)
Hence,
d x d y = | 1 0 v d g ( u ) d u g ( u ) | = g ( u ) . d x d y = 1      0 v d g ( u ) d u g ( u ) = g ( u ) . dx^^dy=|[1,0],[v(dg(u))/(du)g(u)]|=g(u).d x \wedge d y=\left|\begin{array}{rr} 1 & 0 \\ v \frac{d g(u)}{d u} g(u) \end{array}\right|=g(u) .dxdy=|10vdg(u)dug(u)|=g(u).
We conclude with the identity
R f ( x , y ) d y d x = a b 0 1 f ( u , v g ( u ) ) g ( u ) d v d u R f ( x , y ) d y d x = a b 0 1 f ( u , v g ( u ) ) g ( u ) d v d u int_(R)f(x,y)dydx=int_(a)^(b)int_(0)^(1)f(u,vg(u))g(u)dvdu\int_{R} f(x, y) d y d x=\int_{a}^{b} \int_{0}^{1} f(u, v g(u)) g(u) d v d uRf(x,y)dydx=ab01f(u,vg(u))g(u)dvdu
5.17. Let R R RRR be the region below the graph of y = x 2 y = x 2 y=x^(2)y=x^{2}y=x2, and between the lines x = 0 x = 0 x=0x=0x=0 and x = 2 x = 2 x=2x=2x=2. Calculate
R x y 2 d x d y R x y 2 d x d y int_(R)xy^(2)dxdy\int_{R} x y^{2} d x d yRxy2dxdy
A slight variant is to integrate over a region bounded by the graphs of equations y = g 1 ( x ) y = g 1 ( x ) y=g_(1)(x)y=g_{1}(x)y=g1(x) and y = g 2 ( x ) y = g 2 ( x ) y=g_(2)(x)y=g_{2}(x)y=g2(x), and by the lines x = a x = a x=ax=ax=a and x = b x = b x=bx=bx=b, where g 1 ( x ) < g 2 ( x ) g 1 ( x ) < g 2 ( x ) g_(1)(x) < g_(2)(x)g_{1}(x)<g_{2}(x)g1(x)<g2(x) for all x [ a , b ] x [ a , b ] x in[a,b]x \in[a, b]x[a,b]. To compute such an integral we may simply integrate over the region below g 2 ( x ) g 2 ( x ) g_(2)(x)g_{2}(x)g2(x), then integrate over the region below g 1 ( x ) g 1 ( x ) g_(1)(x)g_{1}(x)g1(x), and subtract.
5.18. Let R R RRR be the region to the right of the y y yyy-axis, to the left of the graph of x = g ( y ) x = g ( y ) x=g(y)x=g(y)x=g(y), above the line y = a y = a y=ay=ay=a and below the line y = b y = b y=by=by=b. Find a formula for R f ( x , y ) d x d y R f ( x , y ) d x d y int_(R)f(x,y)dxdy\int_{\mathrm{R}} f(x, y) d x d yRf(x,y)dxdy.
5.19. Let R R RRR be the region in the first quadrant of R 2 R 2 R^(2)\mathrm{R}^{2}R2, below the line y = x y = x y=xy=xy=x, and bounded by x 2 + y 2 = 4 x 2 + y 2 = 4 x^(2)+y^(2)=4x^{2}+y^{2}=4x2+y2=4. Integrate the 2-form over R R RRR.
ω = ( 1 + y 2 x 2 ) d x d y ω = 1 + y 2 x 2 d x d y omega=(1+(y^(2))/(x^(2)))dx^^dy\omega=\left(1+\frac{y^{2}}{x^{2}}\right) d x \wedge d yω=(1+y2x2)dxdy
5.20. Let R R RRR be the region of the x y x y xyx yxy-plane bounded by the ellipse
9 x 2 + 4 y 2 = 36 9 x 2 + 4 y 2 = 36 9x^(2)+4y^(2)=369 x^{2}+4 y^{2}=369x2+4y2=36
Integrate the 2 -form ω = x 2 d x d y ω = x 2 d x d y omega=x^(2)dx^^dy\omega=x^{2} d x \wedge d yω=x2dxdy over R R RRR (Hint. see Problem 2.23 of Chapter 1.)
5.21. Integrate the 2 -form over the top half of the unit sphere using the following parameterization from rectangular coordinates: where
x 2 + y 2 1 x 2 + y 2 1 sqrt()x^(2)+y^(2) <= 1\sqrt{ } x^{2}+y^{2} \leq 1x2+y21. Compare your answer to Problem 5.3.
ω = 1 x d y d z 1 y d x d z ( x , y ) ( x , y , 1 x 2 y 2 ) ω = 1 x d y d z 1 y d x d z ( x , y ) x , y , 1 x 2 y 2 {:[omega=(1)/(x)dy^^dz-(1)/(y)dx^^dz],[(x","y)rarr(x,y,sqrt(1-x^(2)-y^(2)))]:}\begin{aligned} & \omega=\frac{1}{x} d y \wedge d z-\frac{1}{y} d x \wedge d z \\ & (x, y) \rightarrow\left(x, y, \sqrt{1-x^{2}-y^{2}}\right) \end{aligned}ω=1xdydz1ydxdz(x,y)(x,y,1x2y2)

5.6.3 3-forms on 3

Example 28. Let V = { ( r , θ , z ) 1 r 2 , 0 z 1 } V = { ( r , θ , z ) 1 r 2 , 0 z 1 } V={(r,theta,z)∣1 <= r <= 2,0 <= z <= 1}V=\{(r, \theta, z) \mid 1 \leq r \leq 2,0 \leq z \leq 1\}V={(r,θ,z)1r2,0z1}. ( V V VVV is the region between the cylinders of radii one and two and between the planes z z zzz = 0 = 0 =0=0=0 and z = 1 z = 1 z=1z=1z=1.) Let's calculate
V z ( x 2 + y 2 ) d x d y d z V z x 2 + y 2 d x d y d z int_(V)z(x^(2)+y^(2))dx^^dy^^dz\int_{V} z\left(x^{2}+y^{2}\right) d x \wedge d y \wedge d zVz(x2+y2)dxdydz
The region V V VVV is best parameterized using cylindrical coordinates: Ψ ( r , θ , z ) = ( r cos θ , r sin θ , z ) Ψ ( r , θ , z ) = ( r cos θ , r sin θ , z ) Psi(r,theta,z)=(r cos theta,r sin theta,z)\Psi(r, \theta, z)=(r \cos \theta, r \sin \theta, z)Ψ(r,θ,z)=(rcosθ,rsinθ,z),
where 1 r 2 , 1 θ 2 π , and 0 z 1  where  1 r 2 , 1 θ 2 π , and  0 z 1 " where "1 <= r <= 2,1 <= theta <= 2pi", and "0 <= z <= 1\text { where } 1 \leq r \leq 2,1 \leq \theta \leq 2 \pi \text {, and } 0 \leq z \leq 1 where 1r2,1θ2π, and 0z1 We compute the partials:
Ψ r = cos θ , sin θ , 0 Ψ θ = r sin θ , r cos θ , 0 Ψ z = 0 , 0 , 1 . Ψ r = cos θ , sin θ , 0 Ψ θ = r sin θ , r cos θ , 0 Ψ z = 0 , 0 , 1 . {:[(del Psi)/(del r)=(:cos theta","sin theta","0:)],[(del Psi)/(del theta)=(:-r sin theta","r cos theta","0:)],[(del Psi)/(del z)=(:0","0","1:).]:}\begin{aligned} & \frac{\partial \Psi}{\partial r}=\langle\cos \theta, \sin \theta, 0\rangle \\ & \frac{\partial \Psi}{\partial \theta}=\langle-r \sin \theta, r \cos \theta, 0\rangle \\ & \frac{\partial \Psi}{\partial z}=\langle 0,0,1\rangle . \end{aligned}Ψr=cosθ,sinθ,0Ψθ=rsinθ,rcosθ,0Ψz=0,0,1.
Hence,
d x d y d z = | cos θ r sin θ 0 sin θ r cos θ 0 0 0 1 | = r . d x d y d z = cos θ      r sin θ      0 sin θ      r cos θ      0 0      0      1 = r . dx^^dy^^dz=|[cos theta,-r sin theta,0],[sin theta,r cos theta,0],[0,0,1]|=r.d x \wedge d y \wedge d z=\left|\begin{array}{lll} \cos \theta & -r \sin \theta & 0 \\ \sin \theta & r \cos \theta & 0 \\ 0 & 0 & 1 \end{array}\right|=r .dxdydz=|cosθrsinθ0sinθrcosθ0001|=r.
Also,
z ( x 2 + y 2 ) = z ( r 2 cos 2 θ + r 2 sin 2 θ ) = z r 2 . z x 2 + y 2 = z r 2 cos 2 θ + r 2 sin 2 θ = z r 2 . z(x^(2)+y^(2))=z(r^(2)cos^(2)theta+r^(2)sin^(2)theta)=zr^(2).z\left(x^{2}+y^{2}\right)=z\left(r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta\right)=z r^{2} .z(x2+y2)=z(r2cos2θ+r2sin2θ)=zr2.
So we have
V z ( x 2 + y 2 ) d x d y d z = 0 1 0 2 π 1 2 ( z r 2 ) ( r ) d r d θ d z = 0 1 0 2 π 1 2 z r 3 d r d θ d z = 15 4 0 1 0 2 π z d θ d z = 15 π 2 0 1 z d z = 15 π 4 V z x 2 + y 2 d x d y d z = 0 1 0 2 π 1 2 z r 2 ( r ) d r d θ d z = 0 1 0 2 π 1 2 z r 3 d r d θ d z = 15 4 0 1 0 2 π z d θ d z = 15 π 2 0 1 z d z = 15 π 4 {:[int_(V)z(x^(2)+y^(2))dx^^dy^^dz=int_(0)^(1)int_(0)^(2pi)int_(1)^(2)(zr^(2))(r)drd theta dz],[=int_(0)^(1)int_(0)^(2pi)int_(1)^(2)zr^(3)drd theta dz],[=(15)/(4)int_(0)^(1)int_(0)^(2pi)zd theta dz],[=(15 pi)/(2)int_(0)^(1)zdz],[=(15 pi)/(4)]:}\begin{aligned} \int_{V} z\left(x^{2}+y^{2}\right) d x \wedge d y \wedge d z & =\int_{0}^{1} \int_{0}^{2 \pi} \int_{1}^{2}\left(z r^{2}\right)(r) d r d \theta d z \\ & =\int_{0}^{1} \int_{0}^{2 \pi} \int_{1}^{2} z r^{3} d r d \theta d z \\ & =\frac{15}{4} \int_{0}^{1} \int_{0}^{2 \pi} z d \theta d z \\ & =\frac{15 \pi}{2} \int_{0}^{1} z d z \\ & =\frac{15 \pi}{4} \end{aligned}Vz(x2+y2)dxdydz=0102π12(zr2)(r)drdθdz=0102π12zr3drdθdz=1540102πzdθdz=15π201zdz=15π4
5.22. Integrate the 3-form ω = x d x d y d z ω = x d x d y d z omega=xdx^^dy^^dz\omega=x d x \wedge d y \wedge d zω=xdxdydz over the region of R 3 R 3 R^(3)\mathrm{R}^{3}R3 in the first octant bounded by the cylinders x 2 + y 2 = 1 x 2 + y 2 = 1 x^(2)+y^(2)=1x^{2}+y^{2}=1x2+y2=1 and x 2 + y 2 x 2 + y 2 x^(2)+y^(2)x^{2}+y^{2}x2+y2 = 4 = 4 =4=4=4, and the plane z = 2 z = 2 z=2z=2z=2.
5.23. Let R R RRR be the region in the first octant of R 3 R 3 R^(3)R^{3}R3 bounded by the spheres x 2 + y 2 + z 2 = 1 x 2 + y 2 + z 2 = 1 x^(2)+y^(2)+z^(2)=1x^{2}+y^{2}+z^{2}=1x2+y2+z2=1 and x 2 + y 2 + z 2 = 4 x 2 + y 2 + z 2 = 4 x^(2)+y^(2)+z^(2)=4x^{2}+y^{2}+z^{2}=4x2+y2+z2=4. Integrate the 3-form ω = d x d y d z ω = d x d y d z omega=dx^^dy^^dz\omega=d x \wedge d y \wedge d zω=dxdydz over R R RRR.
2 1 + x 2 + y 2 d x d y d z 2 1 + x 2 + y 2 d x d y d z 2sqrt(1+x^(2)+y^(2))dx^^dy^^dz2 \sqrt{1+x^{2}+y^{2}} d x \wedge d y \wedge d z21+x2+y2dxdydz
5.24. Let V V VVV be the volume in the first octant, inside the cylinder of radius one, and below the plane z = 1 z = 1 z=1z=1z=1. Integrate the 3 -form over V V VVV. 5.25. Let V V VVV be the region inside the cylinder of radius one, centered around the z z zzz-axis, and between the planes z = 0 z = 0 z=0z=0z=0 and z = 2 z = 2 z=2z=2z=2. Integrate the function f ( x , y , z ) = z f ( x , y , z ) = z f(x,y,z)=zf(x, y, z)=zf(x,y,z)=z over V V VVV.

5.7 Summary: How to integrate a differential form

5.7.1 The steps

To compute the integral of a differential n n nnn-form, ω ω omega\omegaω, over a region, S S SSS, the steps are as follows:
  1. Choose a parameterization, Ψ : R S Ψ : R S Psi:R rarr S\Psi: R \rightarrow SΨ:RS, where R R RRR is a subset of R n R n R^(n)\mathrm{R}^{n}Rn (see Figure 5.7).
Fig. 5.7.

2. Find all n n nnn vectors given by the partial derivatives of Ψ Ψ Psi\PsiΨ. Geometrically, these are tangent vectors to S S SSS which span its tangent space (see Figure 5.8).
3. Plug the tangent vectors into ω ω omega\omegaω at the point Ψ ( u 1 , u 2 , , u n ) Ψ u 1 , u 2 , , u n Psi(u_(1),u_(2),dots,u_(n))\Psi\left(u_{1}, u_{2}, \ldots, u_{n}\right)Ψ(u1,u2,,un).
4. Integrate the resulting function over R R RRR.
Fig. 5.8.

5.7.2 Integrating 2-forms

The best way to see the above steps in action is to look at the integral of a 2 -form over a surface in R 3 R 3 R^(3)\mathrm{R}^{3}R3. In general, such a 2 -form is given by
ω = f 1 ( x , y , z ) d x d y + f 2 ( x , y , z ) d y d z + f 3 ( x , y , z ) d x d z ω = f 1 ( x , y , z ) d x d y + f 2 ( x , y , z ) d y d z + f 3 ( x , y , z ) d x d z omega=f_(1)(x,y,z)dx^^dy+f_(2)(x,y,z)dy^^dz+f_(3)(x,y,z)dx^^dz\omega=f_{1}(x, y, z) d x \wedge d y+f_{2}(x, y, z) d y \wedge d z+f_{3}(x, y, z) d x \wedge d zω=f1(x,y,z)dxdy+f2(x,y,z)dydz+f3(x,y,z)dxdz.
To integrate ω ω omega\omegaω over S S SSS we now follow the steps:
  1. Choose a parameterization, Ψ : R S Ψ : R S Psi:R rarr S\Psi: R \rightarrow SΨ:RS, where R R RRR is a subset of R 2 R 2 R^(2)\mathrm{R}^{2}R2.
Ψ ( u , v ) = ( g 1 ( u , v ) , g 2 ( u , v ) , g 3 ( u , v ) ) Ψ ( u , v ) = g 1 ( u , v ) , g 2 ( u , v ) , g 3 ( u , v ) Psi(u,v)=(g_(1)(u,v),g_(2)(u,v),g_(3)(u,v))\Psi(u, v)=\left(g_{1}(u, v), g_{2}(u, v), g_{3}(u, v)\right)Ψ(u,v)=(g1(u,v),g2(u,v),g3(u,v))
  1. Find both vectors given by the partial derivatives of Ψ Ψ Psi\PsiΨ.
Ψ u = g 1 u , g 2 u , g 3 u Ψ v = g 1 v , g 2 v , g 3 v . Ψ u = g 1 u , g 2 u , g 3 u Ψ v = g 1 v , g 2 v , g 3 v . {:[(del Psi)/(del u)=(:(delg_(1))/(del u),(delg_(2))/(del u),(delg_(3))/(del u):)],[(del Psi)/(del v)=(:(delg_(1))/(del v),(delg_(2))/(del v),(delg_(3))/(del v):).]:}\begin{aligned} & \frac{\partial \Psi}{\partial u}=\left\langle\frac{\partial g_{1}}{\partial u}, \frac{\partial g_{2}}{\partial u}, \frac{\partial g_{3}}{\partial u}\right\rangle \\ & \frac{\partial \Psi}{\partial v}=\left\langle\frac{\partial g_{1}}{\partial v}, \frac{\partial g_{2}}{\partial v}, \frac{\partial g_{3}}{\partial v}\right\rangle . \end{aligned}Ψu=g1u,g2u,g3uΨv=g1v,g2v,g3v.
  1. Plug the tangent vectors into ω ω omega\omegaω at the point Ψ ( u , U ) Ψ ( u , U ) Psi(u,U)\Psi(u, U)Ψ(u,U). To do this, x , y x , y x,yx, yx,y and z z zzz will come from the coordinates of Ψ Ψ Psi\PsiΨ. That is, x = g 1 ( u , ) x = g 1 ( u , ) x=g_(1)(u,uu)x=g_{1}(u, \cup)x=g1(u,), y = g 2 ( u , U ) y = g 2 ( u , U ) y=g_(2)(u,U)y=g_{2}(u, \mathrm{U})y=g2(u,U) and z = g 3 ( u , U ) z = g 3 ( u , U ) z=g_(3)(u,U)z=g_{3}(u, \mathrm{U})z=g3(u,U). Terms like d x d y d x d y dx^^dyd x \wedge d ydxdy are determinants of 2 × 2 2 × 2 2xx22 \times 22×2 matrices, whose entries come from the vectors computed in the previous step. Geometrically, the value of d x d y d x d y dx^^dyd x \wedge d ydxdy is the area of the parallelogram spanned by the vectors Ψ u Ψ u (del Psi)/(del u)\frac{\partial \Psi}{\partial u}Ψu and Ψ v Ψ v (del Psi)/(del v)\frac{\partial \Psi}{\partial v}Ψv (tangent vectors to S S SSS ), projected onto the d x d y d x d y dxdyd x d ydxdy-plane (see Figure 5.9).
    The result of all this is:
    Fig. 5.9. Evaluating d x d y d x d y dx^^dyd x \wedge d ydxdy geometrically.
f 1 ( g 1 , g 2 , g 3 ) | g 1 u g 1 v g 2 u g 2 v | + f 2 ( g 1 , g 2 , g 3 ) | g 2 u g 2 v g 3 u g 3 v | + f 3 ( g 1 , g 2 , g 3 ) | g 1 u g 1 v g 3 u g 3 v | f 1 g 1 , g 2 , g 3 g 1 u g 1 v g 2 u g 2 v + f 2 g 1 , g 2 , g 3 g 2 u g 2 v g 3 u g 3 v + f 3 g 1 , g 2 , g 3 g 1 u g 1 v g 3 u g 3 v {:[f_(1)(g_(1),g_(2),g_(3))|[(delg_(1))/(del u),(delg_(1))/(del v)],[(delg_(2))/(del u),(delg_(2))/(del v)]|+f_(2)(g_(1),g_(2),g_(3))|[(delg_(2))/(del u),(delg_(2))/(del v)],[(delg_(3))/(del u),(delg_(3))/(del v)]|],[+f_(3)(g_(1),g_(2),g_(3))|[(delg_(1))/(del u)(delg_(1))/(del v)],[(delg_(3))/(del u)(delg_(3))/(del v)]|]:}\begin{gathered} f_{1}\left(g_{1}, g_{2}, g_{3}\right)\left|\begin{array}{ll} \frac{\partial g_{1}}{\partial u} & \frac{\partial g_{1}}{\partial v} \\ \frac{\partial g_{2}}{\partial u} & \frac{\partial g_{2}}{\partial v} \end{array}\right|+f_{2}\left(g_{1}, g_{2}, g_{3}\right)\left|\begin{array}{ll} \frac{\partial g_{2}}{\partial u} & \frac{\partial g_{2}}{\partial v} \\ \frac{\partial g_{3}}{\partial u} & \frac{\partial g_{3}}{\partial v} \end{array}\right| \\ +f_{3}\left(g_{1}, g_{2}, g_{3}\right)\left|\begin{array}{l} \frac{\partial g_{1}}{\partial u} \frac{\partial g_{1}}{\partial v} \\ \frac{\partial g_{3}}{\partial u} \frac{\partial g_{3}}{\partial v} \end{array}\right| \end{gathered}f1(g1,g2,g3)|g1ug1vg2ug2v|+f2(g1,g2,g3)|g2ug2vg3ug3v|+f3(g1,g2,g3)|g1ug1vg3ug3v|
Note that simplifying this gives a function of u u uuu and U U UUU.
4. Integrate the resulting function over R R RRR. In other words, if h ( u , U ) h ( u , U ) h(u,U)h(u, U)h(u,U) is the function you ended up with in the previous step, then compute
R h ( u , v ) d u d v . R h ( u , v ) d u d v . int_(R)h(u,v)dudv.\int_{R} h(u, v) d u d v .Rh(u,v)dudv.
If R R RRR is not a rectangle you may have to find a parameterization of R R RRR whose domain is a rectangle and repeat the above steps to compute this integral.

5.7.3 A sample 2-form

Let ω = ( x 2 + y 2 ) d x d y + z d y d z ω = x 2 + y 2 d x d y + z d y d z omega=(x^(2)+y^(2))dx^^dy+zdy^^dz\omega=\left(x^{2}+y^{2}\right) d x \wedge d y+z d y \wedge d zω=(x2+y2)dxdy+zdydz. Let S S SSS denote the subset of the cylinder x 2 + y = 1 x 2 + y = 1 x^(2)+y=1x^{2}+y=1x2+y=1 that lies between the planes z = 0 z = 0 z=0z=0z=0 and z = 1 z = 1 z=1z=1z=1.
  1. Choose a parameterization, Ψ : R S Ψ : R S Psi:R rarr S\Psi: R \rightarrow SΨ:RS.
Ψ ( θ , z ) = ( cos θ , sin θ , z ) . Ψ ( θ , z ) = ( cos θ , sin θ , z ) . Psi(theta,z)=(cos theta,sin theta,z).\Psi(\theta, z)=(\cos \theta, \sin \theta, z) .Ψ(θ,z)=(cosθ,sinθ,z).
Where R = { ( θ , z ) 0 θ 2 π , 0 z 1 } R = { ( θ , z ) 0 θ 2 π , 0 z 1 } R={(theta,z)∣0 <= theta <= 2pi,0 <= z <= 1}R=\{(\theta, z) \mid 0 \leq \theta \leq 2 \pi, 0 \leq z \leq 1\}R={(θ,z)0θ2π,0z1}.
2. Find both vectors given by the partial derivatives of Ψ Ψ Psi\PsiΨ.
Ψ θ = sin θ , cos θ , 0 Ψ z = 0 , 0 , 1 . Ψ θ = sin θ , cos θ , 0 Ψ z = 0 , 0 , 1 . {:[(del Psi)/(del theta)=(:-sin theta","cos theta","0:)],[(del Psi)/(del z)=(:0","0","1:).]:}\begin{aligned} & \frac{\partial \Psi}{\partial \theta}=\langle-\sin \theta, \cos \theta, 0\rangle \\ & \frac{\partial \Psi}{\partial z}=\langle 0,0,1\rangle . \end{aligned}Ψθ=sinθ,cosθ,0Ψz=0,0,1.
  1. Plug the tangent vectors into ω ω omega\omegaω at the point Ψ ( θ , z ) Ψ ( θ , z ) Psi(theta,z)\Psi(\theta, z)Ψ(θ,z). We get
( cos 2 θ + sin 2 θ ) | sin θ 0 cos θ 0 | + z | cos θ 0 0 1 | . cos 2 θ + sin 2 θ sin θ      0 cos θ      0 + z cos θ 0 0 1 . (cos^(2)theta+sin^(2)theta)|[-sin theta,0],[cos theta,0]|+z|[cos theta,0],[0,1]|.\left(\cos ^{2} \theta+\sin ^{2} \theta\right)\left|\begin{array}{rr} -\sin \theta & 0 \\ \cos \theta & 0 \end{array}\right|+z\left|\begin{array}{cc} \cos \theta & 0 \\ 0 & 1 \end{array}\right| .(cos2θ+sin2θ)|sinθ0cosθ0|+z|cosθ001|.
This simplifies to the function z cos θ z cos θ z cos thetaz \cos \thetazcosθ.
4. Integrate the resulting function over R R RRR.
0 1 0 2 π z cos θ d θ d z 0 1 0 2 π z cos θ d θ d z int_(0)^(1)int_(0)^(2pi)z cos theta d theta dz\int_{0}^{1} \int_{0}^{2 \pi} z \cos \theta d \theta d z0102πzcosθdθdz
Note that the integrand comes from Step 3 and the limits of integration come from Step 1.
6

Differentiation of Forms

6.1 The derivative of a differential 1-form

The goal of this section is to figure out what we mean by the derivative of a differential form. One way to think about a derivative is as a function which measures the variation of some other function. Suppose ω ω omega\omegaω is a 1 -form on R 2 R 2 R^(2)R^{2}R2. What do we mean by the "variation" of ω ω omega\omegaω ? One thing we can try is to plug in a vector field V V VVV. The result is a function from R 2 R 2 R^(2)\mathrm{R}^{2}R2 to R . We can then think about how this function varies near a point p p ppp of R 2 R 2 R^(2)\mathrm{R}^{2}R2. But p p ppp can vary in a lot of ways, so we need to pick one. In Section 1.5, we learned how to take another vector, W W WWW, and use it to vary p p ppp. Hence, the derivative of ω ω omega\omegaω, which we shall denote " a ω a ω a omegaa \omegaaω," is a function that acts on both V V VVV and W W WWW. In other words, it must be a 2 -form!
Let's recall how to vary a function f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) in the direction of a vector W W WWW at a point p p ppp. This was precisely the definition of the directional derivative:
W f ( p ) = f ( p ) W , W f ( p ) = f ( p ) W , grad_(W)f(p)=grad f(p)*W,\nabla_{W} f(p)=\nabla f(p) \cdot W,Wf(p)=f(p)W,
where f ( p ) f ( p ) grad f(p)\nabla f(p)f(p) is the gradient of f f fff at p p ppp :
f ( p ) = f x ( p ) , f y ( p ) . f ( p ) = f x ( p ) , f y ( p ) . grad f(p)=(:(del f)/(del x)(p),(del f)/(del y)(p):).\nabla f(p)=\left\langle\frac{\partial f}{\partial x}(p), \frac{\partial f}{\partial y}(p)\right\rangle .f(p)=fx(p),fy(p).
Going back to the 1 -form ω ω omega\omegaω and the vector field V V VVV, we take the directional derivative of the function ω ( V ) ω ( V ) omega(V)\omega(V)ω(V). Let's do this now for a
specific example. Suppose ω = y d x x 2 d y , V = 1 , 2 , w = 2 , 3 ω = y d x x 2 d y , V = 1 , 2 , w = 2 , 3 omega=ydx-x^(2)dy,V=(:1,2:),w=(:2,3:)\omega=y d x-x^{2} d y, V=\langle 1,2\rangle, w=\langle 2,3\rangleω=ydxx2dy,V=1,2,w=2,3, and p = ( 1 , 1 ) p = ( 1 , 1 ) p=(1,1)p=(1,1)p=(1,1). Then ω ( V ) ω ( V ) omega(V)\omega(V)ω(V) is the function y 2 x 2 y 2 x 2 y-2x^(2)y-2 x^{2}y2x2. Now we compute:
W ω ( V ) = ω ( V ) W = 4 x , 1 2 , 3 = 8 x + 3 W ω ( V ) = ω ( V ) W = 4 x , 1 2 , 3 = 8 x + 3 grad_(W)omega(V)=grad omega(V)*W=(:-4x,1:)*(:2,3:)=-8x+3\nabla_{W} \omega(V)=\nabla \omega(V) \cdot W=\langle-4 x, 1\rangle \cdot\langle 2,3\rangle=-8 x+3Wω(V)=ω(V)W=4x,12,3=8x+3
At the point p = ( 1 , 1 ) p = ( 1 , 1 ) p=(1,1)p=(1,1)p=(1,1) this is the number -5 .
What about the variation of ω ( W ) ω ( W ) omega(W)\omega(W)ω(W), in the direction of V V VVV, at the point p p ppp ? The function ω ( W ) ω ( W ) omega(W)\omega(W)ω(W) is 2 y 3 x 2 2 y 3 x 2 2y-3x^(2)2 y-3 x^{2}2y3x2. We now compute:
V ω ( W ) = ω ( W ) V = 6 x , 2 1 , 2 = 6 x + 4 . V ω ( W ) = ω ( W ) V = 6 x , 2 1 , 2 = 6 x + 4 . grad_(V)omega(W)=grad omega(W)*V=(:-6x,2:)*(:1,2:)=-6x+4.\nabla_{V} \omega(W)=\nabla \omega(W) \cdot V=\langle-6 x, 2\rangle \cdot\langle 1,2\rangle=-6 x+4 .Vω(W)=ω(W)V=6x,21,2=6x+4.
At the point p = ( 1 , 1 ) p = ( 1 , 1 ) p=(1,1)p=(1,1)p=(1,1) this is the number -2 .
This is a small problem. We want d ω d ω d omegad \omegadω, the derivative of ω ω omega\omegaω, to be a 2 -form. Hence, a ω ( V , W ) a ω ( V , W ) a omega(V,W)a \omega(V, W)aω(V,W) should equal a ω ( W , V ) a ω ( W , V ) -a omega(W,V)-a \omega(W, V)aω(W,V). How can we use the variations above to define a ω a ω a omegaa \omegaaω so this is true? Simple. We just define it to be the difference in these variations:
(6.1) d ω ( V , W ) = V ω ( W ) W ω ( V ) . (6.1) d ω ( V , W ) = V ω ( W ) W ω ( V ) . {:(6.1)d omega(V","W)=grad_(V)omega(W)-grad_(W)omega(V).:}\begin{equation*} d \omega(V, W)=\nabla_{V} \omega(W)-\nabla_{W} \omega(V) . \tag{6.1} \end{equation*}(6.1)dω(V,W)=Vω(W)Wω(V).
Hence, in the above example, a ω ( 1 , 2 , 2 , 3 ) a ω ( 1 , 2 , 2 , 3 ) a omega((:1,2:),(:2,3:))a \omega(\langle 1,2\rangle,\langle 2,3\rangle)aω(1,2,2,3), at the point p = p = p=p=p= ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1), is the number 2 ( 5 ) = 3 2 ( 5 ) = 3 -2-(-5)=3-2-(-5)=32(5)=3.
6.1. Suppose w = x y 2 d x + x 3 z d y ( y + z 9 ) d z , V = 1 , 2 , 3 w = x y 2 d x + x 3 z d y y + z 9 d z , V = 1 , 2 , 3 w=xy^(2)dx+x^(3)zdy-(y+z^(9))dz,V=(:1,2,3:)w=x y^{2} d x+x^{3} z d y-\left(y+z^{9}\right) d z, V=\langle 1,2,3\ranglew=xy2dx+x3zdy(y+z9)dz,V=1,2,3, and W W WWW = 1 , 0 , 1 = 1 , 0 , 1 =(:-1,0,1:)=\langle-1,0,1\rangle=1,0,1.
  1. Compute ν ω ( W ) ν ω ( W ) grad_(nu)omega(W)\nabla_{\nu} \omega(W)νω(W) and h ω ( V ) h ω ( V ) grad_(h)omega(V)\nabla_{h} \omega(V)hω(V), at the point (2, 3, -1).
  2. Use your answer to the previous question to compute a ω ( V a ω ( V a omega(Va \omega(Vaω(V, W W WWW ) at the point ( 2 , 3 , 1 ) ( 2 , 3 , 1 ) (2,3,-1)(2,3,-1)(2,3,1).
There are other ways to determine what a ω a ω a omegaa \omegaaω is rather than using Equation 6.1. Recall that a 2 -form acts on a pair of vectors by projecting them onto each coordinate plane, calculating the area
they span, multiplying by some constant, and adding. So the 2 -form is completely determined by the constants that you multiply by after projecting. In order to figure out what these constants are, we are free to examine the action of the 2 -form on any pair of vectors. For example, suppose we have two vectors that lie in the x y x y xyx yxy-plane and span a parallelogram with area one. If we run these through some 2 -form and end up with the number five, then we know that the multiplicative constant for that 2 -form, associated with the x y x y xyx yxy-plane is 5 . This, in turn, tells us that the 2 -form equals 5 d x d y + v 5 d x d y + v 5dx^^dy+v5 d x \wedge d y+\mathrm{v}5dxdy+v. To figure out what v v vvv is, we can examine the action of the 2 -form on other pairs of vectors.
Let's try this with a general differential 2-form on R 3 R 3 R^(3)\mathrm{R}^{3}R3. Such a form always looks like a ω = a ( x , y , z ) d x d y + b ( x , y , z ) d y d z + c ( x , y a ω = a ( x , y , z ) d x d y + b ( x , y , z ) d y d z + c ( x , y a omega=a(x,y,z)dx^^dy+b(x,y,z)dy^^dz+c(x,ya \omega=a(x, y, z) d x \wedge d y+b(x, y, z) d y \wedge d z+c(x, yaω=a(x,y,z)dxdy+b(x,y,z)dydz+c(x,y, z ) d x d z z ) d x d z z)dx^^dzz) d x \wedge d zz)dxdz. To figure out what a ( x , y , z ) a ( x , y , z ) a(x,y,z)a(x, y, z)a(x,y,z) is, for example, all we need to do is determine what a ω a ω a omegaa \omegaaω does to the vectors 1 , 0 , 0 ( x , y , z ) 1 , 0 , 0 ( x , y , z ) (:1,0,0:)_((x,y,z))\langle 1,0,0\rangle_{(x, y, z)}1,0,0(x,y,z) and 0 0 (:0\langle 00, 1 , 0 ( x , y , z ) 1 , 0 ( x , y , z ) 1,0:)_((x,y,z))1,0\rangle_{(x, y, z)}1,0(x,y,z). Let's compute this using Equation 6.1, assuming ω = f ( x ω = f ( x omega=f(x\omega=f(xω=f(x, y , z ) d x + g ( x , y , z ) d y + h ( x , y , z ) d z y , z ) d x + g ( x , y , z ) d y + h ( x , y , z ) d z y,z)dx+g(x,y,z)dy+h(x,y,z)dzy, z) d x+g(x, y, z) d y+h(x, y, z) d zy,z)dx+g(x,y,z)dy+h(x,y,z)dz.
d ω ( 1 , 0 , 0 , 0 , 1 , 0 ) = 1 , 0 , 0 ω ( 0 , 1 , 0 ) 0 , 1 , 0 ω ( 1 , 0 , 0 ) = g x , g y , g z 1 , 0 , 0 f x , f y , f z 0 , 1 , 0 = g x f y . d ω ( 1 , 0 , 0 , 0 , 1 , 0 ) = 1 , 0 , 0 ω ( 0 , 1 , 0 ) 0 , 1 , 0 ω ( 1 , 0 , 0 ) = g x , g y , g z 1 , 0 , 0 f x , f y , f z 0 , 1 , 0 = g x f y . {:[d omega((:1","0","0:)","(:0","1","0:))=grad_((:1,0,0:))omega((:0","1","0:))-grad_((:0,1,0:))omega((:1","0","0:))],[=(:(del g)/(del x),(del g)/(del y),(del g)/(del z):)*(:1","0","0:)-(:(del f)/(del x),(del f)/(del y),(del f)/(del z):)*(:0","1","0:)],[=(del g)/(del x)-(del f)/(del y).]:}\begin{aligned} d \omega(\langle 1,0,0\rangle,\langle 0,1,0\rangle) & =\nabla_{\langle 1,0,0\rangle} \omega(\langle 0,1,0\rangle)-\nabla_{\langle 0,1,0\rangle} \omega(\langle 1,0,0\rangle) \\ & =\left\langle\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z}\right\rangle \cdot\langle 1,0,0\rangle-\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle \cdot\langle 0,1,0\rangle \\ & =\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y} . \end{aligned}dω(1,0,0,0,1,0)=1,0,0ω(0,1,0)0,1,0ω(1,0,0)=gx,gy,gz1,0,0fx,fy,fz0,1,0=gxfy.
Similarly, direct computation shows:
d ω ( 0 , 1 , 0 , ( 0 , 0 , 1 ) = h y g z d ω ( 0 , 1 , 0 , ( 0 , 0 , 1 ) = h y g z d omega((:0,1,0:),(0,0,1:))=(del h)/(del y)-(del g)/(del z)d \omega(\langle 0,1,0\rangle,(0,0,1\rangle)=\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}dω(0,1,0,(0,0,1)=hygz
and,
d ω ( 1 , 0 , 0 , 0 , 0 , 1 ) = h x f z . d ω ( 1 , 0 , 0 , 0 , 0 , 1 ) = h x f z . d omega((:1,0,0:),(:0,0,1:))=(del h)/(del x)-(del f)/(del z).d \omega(\langle 1,0,0\rangle,\langle 0,0,1\rangle)=\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z} .dω(1,0,0,0,0,1)=hxfz.
Hence, we conclude that
d ω = ( g x f y ) d x d y + ( h y g z ) d y d z + ( h x f z ) d x d z . d ω = g x f y d x d y + h y g z d y d z + h x f z d x d z . d omega=((del g)/(del x)-(del f)/(del y))dx^^dy+((del h)/(del y)-(del g)/(del z))dy^^dz+((del h)/(del x)-(del f)/(del z))dx^^dz.d \omega=\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right) d x \wedge d y+\left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right) d y \wedge d z+\left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right) d x \wedge d z .dω=(gxfy)dxdy+(hygz)dydz+(hxfz)dxdz.
6.2. Suppose ω = f ( x , y ) d x + g ( x , y ) d y ω = f ( x , y ) d x + g ( x , y ) d y omega=f(x,y)dx+g(x,y)dy\omega=f(x, y) d x+g(x, y) d yω=f(x,y)dx+g(x,y)dy is a 1-form on R 2 R 2 R^(2)\mathrm{R}^{2}R2. Show that a ω = ( g x f y ) d x d y a ω = g x f y d x d y a omega=((del g)/(del x)-(del f)/(del y))dx^^dya \omega=\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right) d x \wedge d yaω=(gxfy)dxdy.
6.3. If ω = y d x x 2 d y ω = y d x x 2 d y omega=ydx-x^(2)dy\omega=y d x-x^{2} d yω=ydxx2dy, find a ω a ω a omegaa \omegaaω. Verify that a ω ( 1 , 2 , 2 , 3 ) = 3 a ω ( 1 , 2 , 2 , 3 ) = 3 a omega((:1,2:),(:2,3:))=3a \omega(\langle 1,2\rangle,\langle 2,3\rangle)=3aω(1,2,2,3)=3 at the point ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1).
Technical Note: Equation 6.1 defines the value of a w a w awa waw as long as the vector fields V V VVV and W W WWW are constant. If non-constant vector fields are used, then the answer provided by Equation 6.1 will involve partial derivatives of the components of V V VVV and W W WWW, and hence will not be a differential form. Despite this Equation 6.1 does lead to the correct formulas for aw, as in Exercise 6.2 above. Once such formulas are obtained then any vector fields can be plugged in.

6.2 Derivatives of n n n\boldsymbol{n}n-forms

Before jumping to the general case, let's look at the derivative of a 2 -form. A 2 -form, ω ω omega\omegaω, acts on a pair of vector fields, V V VVV and W W WWW, to return a function. To find a variation of ω ω omega\omegaω we can examine how this function varies in the direction of a third vector, U U UUU, at some point p p ppp. Hence, whatever aw turns out to be, it will be a function of the vectors U , V U , V U,VU, VU,V, and W W WWW at each point p p ppp. So, we would like to define it to be a 3 -form.
Let's start by looking at the variation of ω ( V , W ) ω ( V , W ) omega(V,W)\omega(V, W)ω(V,W) in the direction of U U UUU. We write this as L ω ( V , W ) L ω ( V , W ) grad_(L)omega(V,W)\nabla_{\mathcal{L}} \omega(V, W)Lω(V,W). If we were to define this as the value of a ω ( U , V , W ) a ω ( U , V , W ) a omega(U,V,W)a \omega(U, V, W)aω(U,V,W), we would find that, in general, it would not be alternating. That is, usually l ω ( V , W ) l ω ( U , W ) l ω ( V , W ) l ω ( U , W ) grad_(l)omega(V,W)!=-grad_(l)omega(U,W)\nabla_{l} \omega(V, W) \neq-\nabla_{l} \omega(U, W)lω(V,W)lω(U,W). To remedy this, we simply define a ω a ω a omegaa \omegaaω to be the alternating sum of all the variations:
d ω ( U , V , W ) = U ω ( V , W ) V ω ( U , W ) + W ω ( U , V ) . d ω ( U , V , W ) = U ω ( V , W ) V ω ( U , W ) + W ω ( U , V ) . d omega(U,V,W)=grad_(U)omega(V,W)-grad_(V)omega(U,W)+grad_(W)omega(U,V).d \omega(U, V, W)=\nabla_{U} \omega(V, W)-\nabla_{V} \omega(U, W)+\nabla_{W} \omega(U, V) .dω(U,V,W)=Uω(V,W)Vω(U,W)+Wω(U,V).
We leave it to the reader to check that a ω a ω a omegaa \omegaaω is alternating and multilinear (assuming U , V U , V U,VU, VU,V, and W W WWW are constant vector fields).
It should not be hard for the reader to now jump to the general case. Suppose ω ω omega\omegaω is an n n nnn-form and V 1 , , V n + 1 V 1 , , V n + 1 V^(1),dots,V^(n+1)V^{1}, \ldots, V^{n+1}V1,,Vn+1 are n + 1 n + 1 n+1n+1n+1 vector fields. Then we define
d ω ( V 1 , , V n + 1 ) = i = 1 n + 1 ( 1 ) i + 1 V i ω ( V 1 , , V i 1 , V i + 1 , , V n + 1 ) . d ω V 1 , , V n + 1 = i = 1 n + 1 ( 1 ) i + 1 V i ω V 1 , , V i 1 , V i + 1 , , V n + 1 . d omega(V^(1),dots,V^(n+1))=sum_(i=1)^(n+1)(-1)^(i+1)grad_(V^(i))omega(V^(1),dots,V^(i-1),V^(i+1),dots,V^(n+1)).d \omega\left(V^{1}, \ldots, V^{n+1}\right)=\sum_{i=1}^{n+1}(-1)^{i+1} \nabla_{V^{i}} \omega\left(V^{1}, \ldots, V^{i-1}, V^{i+1}, \ldots, V^{n+1}\right) .dω(V1,,Vn+1)=i=1n+1(1)i+1Viω(V1,,Vi1,Vi+1,,Vn+1).
In other words, a ω a ω a omegaa \omegaaω, applied to n + 1 n + 1 n+1n+1n+1 vectors, is the alternating sum of the variations of ω ω omega\omegaω applied to n n nnn of those vectors in the direction of the remaining one. Note that we can think of " d d ddd " as an operator which takes n n nnn-forms to ( n + 1 n + 1 n+1n+1n+1 )-forms.
6.4. Show that d w d w dwd wdw is alternating.
6.5. Show that d ( w + v ) = a w + a v d ( w + v ) = a w + a v d(w+v)=aw+avd(w+v)=a w+a vd(w+v)=aw+av and d ( c ω ) = c a w d ( c ω ) = c a w d(c omega)=cawd(c \omega)=c a wd(cω)=caw, for any constant c c ccc.
6.6. Suppose ω = f ( x , y , z ) d x d y + g ( x , y , z ) d y d z + h ( x , y , z ) ω = f ( x , y , z ) d x d y + g ( x , y , z ) d y d z + h ( x , y , z ) omega=f(x,y,z)dx^^dy+g(x,y,z)dy^^dz+h(x,y,z)\omega=f(x, y, z) d x \wedge d y+g(x, y, z) d y \wedge d z+h(x, y, z)ω=f(x,y,z)dxdy+g(x,y,z)dydz+h(x,y,z) d x d z d x d z dx^^dzd x \wedge d zdxdz. Find a w a w awa waw (Hint. Compute a w ( 1 , 0 , 0 , 0 , 1 , 0 , 0 , 0 , 1 ) a w ( 1 , 0 , 0 , 0 , 1 , 0 , 0 , 0 , 1 ) aw((:1,0,0:),(:0,1,0:),(:0,0,1:))a w(\langle 1,0,0\rangle,\langle 0,1,0\rangle,\langle 0,0,1\rangle)aw(1,0,0,0,1,0,0,0,1) ). Compute d ( x 2 y d x d y + y 2 z d y d z ) d x 2 y d x d y + y 2 z d y d z d(x^(2)ydx^^dy+y^(2)zdy^^dz)d\left(x^{2} y d x \wedge d y+y^{2} z d y \wedge d z\right)d(x2ydxdy+y2zdydz).

6.3 Interlude: 0-forms

Let's go back to Section 4.1, when we introduced coordinates for vectors. At that time, we noted that if C C CCC was the graph of the function y = f ( x ) y = f ( x ) y=f(x)y=f(x)y=f(x) and p p ppp was a point of C C CCC, then the tangent line to C C CCC at p p ppp lies in T p R 2 T p R 2 T_(p)R^(2)T_{p} \mathrm{R}^{2}TpR2 and has equation d y = m d x d y = m d x dy=mdxd y=m d xdy=mdx, for some constant, m m mmm. Of course, if p = ( x 0 , y 0 ) p = x 0 , y 0 p=(x_(0),y_(0))p=\left(x_{0}, y_{0}\right)p=(x0,y0), then m m mmm is just the derivative of f f fff evaluated at x 0 x 0 x_(0)x_{0}x0.
Now, suppose we had looked at the graph of a function of 2variables, z = f ( x , y ) z = f ( x , y ) z=f(x,y)z=f(x, y)z=f(x,y), instead. At some point, p = ( x 0 , y 0 , z 0 ) p = x 0 , y 0 , z 0 p=(x_(0),y_(0),z_(0))p=\left(x_{0}, y_{0}, z_{0}\right)p=(x0,y0,z0), on the graph we could look at the tangent plane, which lies in T p R 3 T p R 3 T_(p)R^(3)T_{p} \mathrm{R}^{3}TpR3. Its equation is d z = m 1 d x + m 2 d y d z = m 1 d x + m 2 d y dz=m_(1)dx+m_(2)dyd z=m_{1} d x+m_{2} d ydz=m1dx+m2dy. Since z = f ( x , y ) , m 1 = f x ( x 0 , y 0 ) z = f ( x , y ) , m 1 = f x x 0 , y 0 ^(z)=f(x,y),m_(1)=(del f)/(del x)(x_(0),y_(0))^{z}=f(x, y), m_{1}=\frac{\partial f}{\partial x}\left(x_{0}, y_{0}\right)z=f(x,y),m1=fx(x0,y0) and m 2 = f y ( x 0 , y 0 ) m 2 = f y x 0 , y 0 ^(m_(2))=(del f)/(del y)(x_(0),y_(0))^{m_{2}}=\frac{\partial f}{\partial y}\left(x_{0}, y_{0}\right)m2=fy(x0,y0), we can rewrite this as
d f = f x d x + f y d y . d f = f x d x + f y d y . df=(del f)/(del x)dx+(del f)/(del y)dy.d f=\frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y .df=fxdx+fydy.
Notice that the right-hand side of this equation is a differential 1form. This is a bit strange; we applied the " d d ddd " operator to something and the result was a 1 -form. However, we know that when we apply the " d d ddd " operator to a differential n n nnn-form we get a differential ( n + 1 ) ( n + 1 ) (n+1)(n+1)(n+1) form. So, it must be that f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) is a differential 0 -form on R 2 R 2 R^(2)\mathrm{R}^{2}R2 !
In retrospect, this should not be so surprising. After all, the input to a differential n n nnn-form on R m R m R^(m)\mathrm{R}^{m}Rm is a point, and n n nnn vectors based at that point. So, the input to a differential 0 -form should be a point of R m R m R^(m)\mathrm{R}^{m}Rm,
and no vectors. In other words, a 0 -form on R m R m R^(m)\mathrm{R}^{m}Rm is just another word for a real-valued function on R m R m R^(m)\mathrm{R}^{m}Rm.
Let's extend some of the things we can do with forms to 0-forms. Suppose f f fff is a 0 -form, and ω ω omega\omegaω is an n n nnn-form (where n n nnn may also be 0 ). What do we mean by f ω f ω f^^omegaf \wedge \omegafω ? Since the wedge product of an n n nnn-form and an m m mmm-form is an ( n + m n + m n+mn+mn+m )-form, it must be that f ω f ω f^^omegaf \wedge \omegafω is an n n nnn form. It is hard to think of any other way to define this as just the product, f ω f ω f omegaf \omegafω.
What about integration? Remember that we integrate n n nnn-forms over subsets of R m R m R^(m)\mathrm{R}^{m}Rm that can be parameterized by a subset of R n R n R^(n)\mathrm{R}^{n}Rn. So 0 -forms get integrated over things parameterized by R 0 R 0 R^(0)\mathrm{R}^{0}R0. In other words, we integrate a 0 -form over a point. How do we do this? We do the simplest possible thing; define the value of a 0 -form, f f fff, integrated over the point, p p ppp, to be ± f ( p ) ± f ( p ) +-f(p)\pm f(p)±f(p). To specify an orientation we just need to say whether or not to use the - sign. We do this just by writing "- p p ppp " instead of " p p ppp " when we want the integral of f f fff over p p ppp to be f ( p ) f ( p ) -f(p)-f(p)f(p).
One word of caution here...beware of orientations! If p R n p R n p inR^(n)p \in \mathrm{R}^{n}pRn, then we use the notation " p p -p-pp " to denote p p ppp with the negative orientation. So if p = 3 R 1 p = 3 R 1 p=-3inR^(1)p=-3 \in \mathrm{R}^{1}p=3R1, then p p -p-pp is not the same as the point, 3. p p -p-pp is just the point, -3 , with a negative orientation. So, if f ( x ) = x 2 f ( x ) = x 2 f(x)=x^(2)f(x)=x^{2}f(x)=x2, then p f p f int_(-p)f\int_{-p} fpf = f ( p ) = 9 = f ( p ) = 9 =-f(p)=-9=-f(p)=-9=f(p)=9.
6.7. If f f fff is the 0 -form x 2 y 3 , p x 2 y 3 , p x^(2)y^(3),px^{2} y^{3}, px2y3,p is the point ( 1 , 1 ) , q ( 1 , 1 ) , q (-1,1),q(-1,1), q(1,1),q is the point ( 1 , 1 ) 1 ) -1)-1)1), and r r rrr is the point ( 1 , 1 1 , 1 -1,-1-1,-11,1 ), then compute the integral of f f fff over the points p , q p , q -p,-q-p,-qp,q, and r r -r-rr, with the indicated orientations.
Let's go back to our exploration of derivatives of n n nnn-forms. Suppose f ( x , y ) d x f ( x , y ) d x f(x,y)dxf(x, y) d xf(x,y)dx is a 1 -form on R 2 R 2 R^(2)\mathrm{R}^{2}R2. Then we have already shown that d ( f d x ) = f y d y d x d ( f d x ) = f y d y d x d(fdx)=(del f)/(del y)dy^^dxd(f d x)=\frac{\partial f}{\partial y} d y \wedge d xd(fdx)=fydydx. We now compute:
d f d x = ( f x d x + f y d y ) d x = f x d x d x + f y d y d x = f y d y d x = d ( f d x ) . d f d x = f x d x + f y d y d x = f x d x d x + f y d y d x = f y d y d x = d ( f d x ) . {:[df^^dx=((del f)/(del x)dx+(del f)/(del y)dy)^^dx],[=(del f)/(del x)dx^^dx+(del f)/(del y)dy^^dx],[=(del f)/(del y)dy^^dx],[=d(fdx).]:}\begin{aligned} d f \wedge d x & =\left(\frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y\right) \wedge d x \\ & =\frac{\partial f}{\partial x} d x \wedge d x+\frac{\partial f}{\partial y} d y \wedge d x \\ & =\frac{\partial f}{\partial y} d y \wedge d x \\ & =d(f d x) . \end{aligned}dfdx=(fxdx+fydy)dx=fxdxdx+fydydx=fydydx=d(fdx).
6.8. If f f fff is a 0 -form, show that d ( f d x 1 d x 2 d x n ) = d f d x 1 d f d x 1 d x 2 d x n = d f d x 1 d(fdx_(1)^^dx_(2)^^dots^^dx_(n))=df^^dx_(1)d\left(f d x_{1} \wedge d x_{2} \wedge \ldots \wedge d x_{n}\right)=d f \wedge d x_{1}d(fdx1dx2dxn)=dfdx1 d x 2 d x n d x 2 d x n ^^dx_(2)^^dots^^dx_(n)\wedge d x_{2} \wedge \ldots \wedge d x_{n}dx2dxn.
6.9. Prove: d ( a ω ) = 0 d ( a ω ) = 0 d(a omega)=0d(a \omega)=0d(aω)=0.
6.10. If ω ω omega\omegaω is an n n nnn-form, and μ μ mu\muμ is an m m mmm-form, then show that d ( ω d ( ω d(omega^^d(\omega \wedged(ω μ ) = a ω μ + ( 1 ) n ω d μ μ ) = a ω μ + ( 1 ) n ω d μ mu)=a omega^^mu+(-1)^(n)omega^^d mu\mu)=a \omega \wedge \mu+(-1)^{n} \omega \wedge d \muμ)=aωμ+(1)nωdμ.

6.4 Algebraic computation of derivatives

As in Section 4.7 we break with the spirit of the text to list the identities we have acquired, and work a few examples.
Let ω ω omega\omegaω be an n n nnn-form, μ μ mu\muμ an m m mmm-form, and f f fff a 0 -form. Then we have the following identities:
d ( d ω ) = 0 d ( ω + μ ) = d ω + d μ d ( ω μ ) = d ω μ + ( 1 ) n ω d μ d ( f d x 1 d x 2 d x n ) = d f d x 1 d x 2 d x n d f = f x 1 d x 1 + f x 2 d x 2 + + f x n d x n . d ( d ω ) = 0 d ( ω + μ ) = d ω + d μ d ( ω μ ) = d ω μ + ( 1 ) n ω d μ d f d x 1 d x 2 d x n = d f d x 1 d x 2 d x n d f = f x 1 d x 1 + f x 2 d x 2 + + f x n d x n . {:[d(d omega)=0],[d(omega+mu)=d omega+d mu],[d(omega^^mu)=d omega^^mu+(-1)^(n)omega^^d mu],[d(fdx_(1)^^dx_(2)^^dots^^dx_(n))=df^^dx_(1)^^dx_(2)^^dots^^dx_(n)],[df=(del f)/(delx_(1))dx_(1)+(del f)/(delx_(2))dx_(2)+dots+(del f)/(delx_(n))dx_(n).]:}\begin{aligned} d(d \omega) & =0 \\ d(\omega+\mu) & =d \omega+d \mu \\ d(\omega \wedge \mu) & =d \omega \wedge \mu+(-1)^{n} \omega \wedge d \mu \\ d\left(f d x_{1} \wedge d x_{2} \wedge \ldots \wedge d x_{n}\right) & =d f \wedge d x_{1} \wedge d x_{2} \wedge \ldots \wedge d x_{n} \\ d f & =\frac{\partial f}{\partial x_{1}} d x_{1}+\frac{\partial f}{\partial x_{2}} d x_{2}+\ldots+\frac{\partial f}{\partial x_{n}} d x_{n} . \end{aligned}d(dω)=0d(ω+μ)=dω+dμd(ωμ)=dωμ+(1)nωdμd(fdx1dx2dxn)=dfdx1dx2dxndf=fx1dx1+fx2dx2++fxndxn.
Example 29.
d ( x y d x x y d y + x y 2 z 3 d z ) = d ( x y ) d x d ( x y ) d y + d ( x y 2 z 3 ) d z = ( y d x + x d y ) d x ( y d x + x d y ) d y + ( y 2 z 3 d x + 2 x y z 3 d y + 3 x y 2 z 2 d z ) d z = y d x d x + x d y d x y d x d y x d y d y + y 2 z 3 d x d z + 2 x y z 3 d y d z + 3 x y 2 z 2 d z d z = x d y d x y d x d y + y 2 z 3 d x d z + 2 x y z 3 d y d z = x d x d y y d x d y + y 2 z 3 d x d z + 2 x y z 3 d y d z = ( x y ) d x d y + y 2 z 3 d x d z + 2 x y z 3 d y d z d ( x y d x x y d y + x y 2 z 3 d z = d ( x y ) d x d ( x y ) d y + d x y 2 z 3 d z = ( y d x + x d y ) d x ( y d x + x d y ) d y + y 2 z 3 d x + 2 x y z 3 d y + 3 x y 2 z 2 d z d z = y d x d x + x d y d x y d x d y x d y d y + y 2 z 3 d x d z + 2 x y z 3 d y d z + 3 x y 2 z 2 d z d z = x d y d x y d x d y + y 2 z 3 d x d z + 2 x y z 3 d y d z = x d x d y y d x d y + y 2 z 3 d x d z + 2 x y z 3 d y d z = ( x y ) d x d y + y 2 z 3 d x d z + 2 x y z 3 d y d z {:[d(xydx-{:xydy+xy^(2)z^(3)dz)],[=d(xy)^^dx-d(xy)^^dy+d(xy^(2)z^(3))^^dz],[=(ydx+xdy)^^dx-(ydx+xdy)^^dy],[+(y^(2)z^(3)dx+2xyz^(3)dy+3xy^(2)z^(2)dz)^^dz],[=ydx^^dx+xdy^^dx-ydx^^dy-xdy^^dy],[+y^(2)z^(3)dx^^dz+2xyz^(3)dy^^dz+3xy^(2)z^(2)dz^^dz],[=xdy^^dx-ydx^^dy+y^(2)z^(3)dx^^dz+2xyz^(3)dy^^dz],[=-xdx^^dy-ydx^^dy+y^(2)z^(3)dx^^dz+2xyz^(3)dy^^dz],[=(-x-y)dx^^dy+y^(2)z^(3)dx^^dz+2xyz^(3)dy^^dz]:}\begin{aligned} d(x y d x- & \left.x y d y+x y^{2} z^{3} d z\right) \\ = & d(x y) \wedge d x-d(x y) \wedge d y+d\left(x y^{2} z^{3}\right) \wedge d z \\ = & (y d x+x d y) \wedge d x-(y d x+x d y) \wedge d y \\ & +\left(y^{2} z^{3} d x+2 x y z^{3} d y+3 x y^{2} z^{2} d z\right) \wedge d z \\ = & y d x \wedge d x+x d y \wedge d x-y d x \wedge d y-x d y \wedge d y \\ & +y^{2} z^{3} d x \wedge d z+2 x y z^{3} d y \wedge d z+3 x y^{2} z^{2} d z \wedge d z \\ = & x d y \wedge d x-y d x \wedge d y+y^{2} z^{3} d x \wedge d z+2 x y z^{3} d y \wedge d z \\ = & -x d x \wedge d y-y d x \wedge d y+y^{2} z^{3} d x \wedge d z+2 x y z^{3} d y \wedge d z \\ = & (-x-y) d x \wedge d y+y^{2} z^{3} d x \wedge d z+2 x y z^{3} d y \wedge d z \end{aligned}d(xydxxydy+xy2z3dz)=d(xy)dxd(xy)dy+d(xy2z3)dz=(ydx+xdy)dx(ydx+xdy)dy+(y2z3dx+2xyz3dy+3xy2z2dz)dz=ydxdx+xdydxydxdyxdydy+y2z3dxdz+2xyz3dydz+3xy2z2dzdz=xdydxydxdy+y2z3dxdz+2xyz3dydz=xdxdyydxdy+y2z3dxdz+2xyz3dydz=(xy)dxdy+y2z3dxdz+2xyz3dydz
Example 30.
d ( x 2 ( y + z 2 ) d x d y + z ( x 3 + y ) d y d z ) = d ( x 2 ( y + z 2 ) ) d x d y + d ( z ( x 3 + y ) ) d y d z = 2 x 2 z d z d x d y + 3 x 2 z d x d y d z = 5 x 2 z d x d y d z d x 2 ( y + z 2 d x d y + z x 3 + y d y d z = d x 2 y + z 2 d x d y + d z x 3 + y d y d z = 2 x 2 z d z d x d y + 3 x 2 z d x d y d z = 5 x 2 z d x d y d z {:[d(x^(2)(y:}{:+z^(2))dx^^dy+z(x^(3)+y)dy^^dz)],[=d(x^(2)(y+z^(2)))^^dx^^dy+d(z(x^(3)+y))^^dy^^dz],[=2x^(2)zdz^^dx^^dy+3x^(2)zdx^^dy^^dz],[=5x^(2)zdx^^dy^^dz]:}\begin{aligned} d\left(x^{2}(y\right. & \left.\left.+z^{2}\right) d x \wedge d y+z\left(x^{3}+y\right) d y \wedge d z\right) \\ & =d\left(x^{2}\left(y+z^{2}\right)\right) \wedge d x \wedge d y+d\left(z\left(x^{3}+y\right)\right) \wedge d y \wedge d z \\ & =2 x^{2} z d z \wedge d x \wedge d y+3 x^{2} z d x \wedge d y \wedge d z \\ & =5 x^{2} z d x \wedge d y \wedge d z \end{aligned}d(x2(y+z2)dxdy+z(x3+y)dydz)=d(x2(y+z2))dxdy+d(z(x3+y))dydz=2x2zdzdxdy+3x2zdxdydz=5x2zdxdydz
6.11. For each differential n n nnn-form, ω ω omega\omegaω, find a ω a ω a omegaa \omegaaω.
  1. sin y d x + cos x d y sin y d x + cos x d y sin ydx+cos xdy\sin y d x+\cos x d ysinydx+cosxdy.
  2. x y 2 d x + x 3 z d y ( y + z 9 ) d z x y 2 d x + x 3 z d y y + z 9 d z xy^(2)dx+x^(3)zdy-(y+z^(9))dzx y^{2} d x+x^{3} z d y-\left(y+z^{9}\right) d zxy2dx+x3zdy(y+z9)dz.
  3. x y 2 d y d z + x 3 z d x d z ( y + z 9 ) d x d y x y 2 d y d z + x 3 z d x d z y + z 9 d x d y xy^(2)dy^^dz+x^(3)zdx^^dz-(y+z^(9))dx^^dyx y^{2} d y \wedge d z+x^{3} z d x \wedge d z-\left(y+z^{9}\right) d x \wedge d yxy2dydz+x3zdxdz(y+z9)dxdy.
  4. x 2 y 3 z 4 d x d y d z x 2 y 3 z 4 d x d y d z x^(2)y^(3)z^(4)dx^^dy^^dzx^{2} y^{3} z^{4} d x \wedge d y \wedge d zx2y3z4dxdydz.
    6.12. If f f fff is the 0 -form x 2 y 3 x 2 y 3 x^(2)y^(3)x^{2} y^{3}x2y3 and ω ω omega\omegaω is the 1 -form x 2 z d x + y 3 z 2 d y x 2 z d x + y 3 z 2 d y x^(2)zdx+y^(3)z^(2)dyx^{2} z d x+y^{3} z^{2} d yx2zdx+y3z2dy (on R 3 R 3 R^(3)\mathrm{R}^{3}R3 ), then use the identity d ( f a ω ) = d f a ω d ( f a ω ) = d f a ω d(fa omega)=df^^a omegad(f a \omega)=d f \wedge a \omegad(faω)=dfaω to compute d ( f a ω d ( f a ω d(fa omegad(f a \omegad(faω ).
    6.13. Let f , g f , g f,gf, gf,g and h h hhh be functions from R 3 R 3 R^(3)\mathrm{R}^{3}R3 to R. If
ω = f d y d z g d x d z + h d x d y ω = f d y d z g d x d z + h d x d y omega=fdy^^dz-gdx^^dz+hdx^^dy\omega=f d y \wedge d z-g d x \wedge d z+h d x \wedge d yω=fdydzgdxdz+hdxdy

6.5 Antiderivatives

Just as in single-variable calculus it will be helpful to have some proficiency in recognizing antiderivatives. Nothing substitutes for practice...
6.14. Find forms whose derivatives are
  1. d x d y d x d y dx^^dyd x \wedge d ydxdy.
  2. d x d y d z d x d y d z dx^^dy^^dzd x \wedge d y \wedge d zdxdydz.
  3. y z d x + x z d y + x y d z y z d x + x z d y + x y d z yzdx+xzdy+xydzy z d x+x z d y+x y d zyzdx+xzdy+xydz.
  4. y 2 z 2 d x + 2 x y z 2 d y + 2 x y 2 z d z y 2 z 2 d x + 2 x y z 2 d y + 2 x y 2 z d z y^(2)z^(2)dx+2xyz^(2)dy+2xy^(2)zdzy^{2} z^{2} d x+2 x y z^{2} d y+2 x y^{2} z d zy2z2dx+2xyz2dy+2xy2zdz.
  5. ( y 2 2 x y ) cos ( x y 2 ) d x d y y 2 2 x y cos x y 2 d x d y (y^(2)-2xy)cos(xy^(2))dx^^dy\left(y^{2}-2 x y\right) \cos \left(x y^{2}\right) d x \wedge d y(y22xy)cos(xy2)dxdy.
    6.15. Show that ω = x y 2 d x ω = x y 2 d x omega=xy^(2)dx\omega=x y^{2} d xω=xy2dx is not the derivative of any 0 -form. (Hint. consider aw.)

7

Stokes' Theorem

7.1 Cells and chains

Up until now, we have not been very specific as to the types of subsets of R m R m R^(m)\mathrm{R}^{m}Rm on which one integrates a differential n n nnn-form. All we have needed is a subset that can be parameterized by a region in R n R n R^(n)R^{n}Rn. To go further we need to specify the types of regions.
Definition 1. Let I = [ 0 , 1 ] I = [ 0 , 1 ] I=[0,1]I=[0,1]I=[0,1]. An n n nnn-cell, σ σ sigma\sigmaσ, is the image of a differentiable map, φ : I n R m φ : I n R m varphi:I^(n)rarrR^(m)\varphi: I^{n} \rightarrow \mathrm{R}^{m}φ:InRm, with a specified orientation. We denote the same cell with opposite orientation as σ σ -sigma-\sigmaσ. We define a 0 cell to be an oriented point of R m R m R^(m)\mathrm{R}^{m}Rm.
Example 31. Suppose g 1 ( x ) g 1 ( x ) g_(1)(x)g_{1}(x)g1(x) and g 2 ( x ) g 2 ( x ) g_(2)(x)g_{2}(x)g2(x) are functions such that g 1 ( x ) < g 1 ( x ) < g_(1)(x) <g_{1}(x)<g1(x)< g 2 ( x ) g 2 ( x ) g_(2)(x)g_{2}(x)g2(x) for all x [ a , b ] x [ a , b ] x in[a,b]x \in[a, b]x[a,b]. Let R R RRR denote the subset of R 2 R 2 R^(2)\mathrm{R}^{2}R2 bounded by the graphs of the equations y = g 1 ( x ) y = g 1 ( x ) y=g_(1)(x)y=g_{1}(x)y=g1(x) and y = g 2 ( x ) y = g 2 ( x ) y=g_(2)(x)y=g_{2}(x)y=g2(x), and by the lines x x xxx = a = a =a=a=a and x = b x = b x=bx=bx=b. In Example 13, we showed that R R RRR is a 2-cell (assuming the induced orientation).
We would like to treat cells as algebraic objects which can be added and subtracted. But if σ σ sigma\sigmaσ is a cell, it may not at all be clear what " 2 σ 2 σ 2sigma2 \sigma2σ " represents. One way to think about it is as two copies of σ σ sigma\sigmaσ, placed right on top of each other.
Definition 2. An n-chain is a formal linear combination of n-cells.
As one would expect, we assume the following relations hold:
σ σ = n σ + m σ = ( n + m ) σ σ + τ = τ + σ σ σ = n σ + m σ = ( n + m ) σ σ + τ = τ + σ {:[sigma-sigma=O/],[n sigma+m sigma=(n+m)sigma],[sigma+tau=tau+sigma]:}\begin{aligned} \sigma-\sigma & =\emptyset \\ n \sigma+m \sigma & =(n+m) \sigma \\ \sigma+\tau & =\tau+\sigma \end{aligned}σσ=nσ+mσ=(n+m)σσ+τ=τ+σ
You may be able to guess what the integral of an n n nnn-form, ω ω omega\omegaω, over an n n nnn-chain is. Suppose C = n σ i C = n σ i C=sum nsigma_(i)C=\sum n \sigma_{i}C=nσi. Then we define
C ω = i n i σ i ω . C ω = i n i σ i ω . int_(C)omega=sum_(i)n_(i)int_(sigma_(i))omega.\int_{C} \omega=\sum_{i} n_{i} \int_{\sigma_{i}} \omega .Cω=iniσiω.
7.1. If f f fff is the 0 -form x 2 y 3 , p x 2 y 3 , p x^(2)y^(3),px^{2} y^{3}, px2y3,p is the point ( 1 , 1 ) , q ( 1 , 1 ) , q (-1,1),q(-1,1), q(1,1),q is the point ( 1 , -1 ), and r r rrr is the point ( 1 , 1 ) ( 1 , 1 ) (-1,-1)(-1,-1)(1,1), then compute the integral of f f fff over the following 0 -chains:
  1. p q ; r p p q ; r p p-q;r-pp-q ; r-ppq;rp.
  2. p + q r p + q r p+q-rp+q-rp+qr.
Another concept that will be useful for us is the boundary of an n n nnn chain. As a warm-up, we define the boundary of a 1 -cell. Suppose σ σ sigma\sigmaσ is the 1 -cell which is the image of φ : [ 0 , 1 ] R m φ : [ 0 , 1 ] R m varphi:[0,1]rarrR^(m)\varphi:[0,1] \rightarrow \mathrm{R}^{m}φ:[0,1]Rm with the induced orientation. Then we define the boundary of σ σ sigma\sigmaσ (which we shall denote " ¯ σ ¯ σ bar(del)sigma\bar{\partial} \sigma¯σ ") as the 0 -chain, φ ( 1 ) φ ( 0 ) φ ( 1 ) φ ( 0 ) varphi(1)-varphi(0)\varphi(1)-\varphi(0)φ(1)φ(0). We can represent this pictorially as in Figure 7.1.
Fig. 7.1. Orienting the boundary of a 1 -cell.
Fig. 7.2. The boundary of a 2-cell.
Figure 7.2 depicts a 2 -cell and its boundary. Notice that the boundary consists of four individually oriented 1 -cells. This hints at the general formula. In general, if the n n nnn-cell σ σ sigma\sigmaσ is the image of the parameterization φ : I n R m φ : I n R m varphi:I^(n)rarrR^(m)\varphi: I^{n} \rightarrow \mathrm{R}^{m}φ:InRm with the induced orientation then
σ = i = 1 n ( 1 ) i + 1 ( ϕ | x 1 , , x i 1 , 1 , x i + 1 , , x n ) ϕ | ( x 1 , , x i 1 , 0 , x i + 1 , , x n ) ) . σ = i = 1 n ( 1 ) i + 1 ϕ x 1 , , x i 1 , 1 , x i + 1 , , x n ϕ x 1 , , x i 1 , 0 , x i + 1 , , x n . del sigma=sum_(i=1)^(n)(-1)^(i+1)( phi|_(x_(1),dots,dots)^({:x_(i-1),1,x_(i+1),dots,x_(n)))- phi|_((x_(1),dots,x_(i-1),0,x_(i+1),dots,x_(n)))).\partial \sigma=\sum_{i=1}^{n}(-1)^{i+1}\left(\left.\phi\right|_{x_{1}, \ldots, \ldots} ^{\left.x_{i-1}, 1, x_{i+1}, \ldots, x_{n}\right)}-\left.\phi\right|_{\left(x_{1}, \ldots, x_{i-1}, 0, x_{i+1}, \ldots, x_{n}\right)}\right) .σ=i=1n(1)i+1(ϕ|x1,,xi1,1,xi+1,,xn)ϕ|(x1,,xi1,0,xi+1,,xn)).
So, if σ σ sigma\sigmaσ is a 2 -cell, then
σ = ( ϕ ( 1 , x 2 ) ϕ ( 0 , x 2 ) ) ( ϕ ( x 1 , 1 ) ϕ ( x 1 , 0 ) ) = ϕ ( 1 , x 2 ) ϕ ( 0 , x 2 ) ϕ ( x 1 , 1 ) + ϕ ( x 1 , 0 ) . σ = ϕ 1 , x 2 ϕ 0 , x 2 ϕ x 1 , 1 ϕ x 1 , 0 = ϕ 1 , x 2 ϕ 0 , x 2 ϕ x 1 , 1 + ϕ x 1 , 0 . {:[del sigma=(phi(1,x_(2))-phi(0,x_(2)))-(phi(x_(1),1)-phi(x_(1),0))],[=phi(1,x_(2))-phi(0,x_(2))-phi(x_(1),1)+phi(x_(1),0).]:}\begin{aligned} \partial \sigma & =\left(\phi\left(1, x_{2}\right)-\phi\left(0, x_{2}\right)\right)-\left(\phi\left(x_{1}, 1\right)-\phi\left(x_{1}, 0\right)\right) \\ & =\phi\left(1, x_{2}\right)-\phi\left(0, x_{2}\right)-\phi\left(x_{1}, 1\right)+\phi\left(x_{1}, 0\right) . \end{aligned}σ=(ϕ(1,x2)ϕ(0,x2))(ϕ(x1,1)ϕ(x1,0))=ϕ(1,x2)ϕ(0,x2)ϕ(x1,1)+ϕ(x1,0).
The four terms on the right side of this equality are the four 1cells depicted in Figure 7.2. The signs in front of these terms guarantee that the orientations are as pictured.
If σ σ sigma\sigmaσ is a 3 -cell, then
σ = ( ϕ ( 1 , x 2 , x 3 ) ϕ ( 0 , x 2 , x 3 ) ) ( ϕ ( x 1 , 1 , x 3 ) ϕ ( x 1 , 0 , x 3 ) ) + ( ϕ ( x 1 , x 2 , 1 ) ϕ ( x 1 , x 2 , 0 ) ) = ϕ ( 1 , x 2 , x 3 ) ϕ ( 0 , x 2 , x 3 ) ϕ ( x 1 , 1 , x 3 ) + ϕ ( x 1 , 0 , x 3 ) + ϕ ( x 1 , x 2 , 1 ) ϕ ( x 1 , x 2 , 0 ) . σ = ϕ 1 , x 2 , x 3 ϕ 0 , x 2 , x 3 ϕ x 1 , 1 , x 3 ϕ x 1 , 0 , x 3 + ϕ x 1 , x 2 , 1 ϕ x 1 , x 2 , 0 = ϕ 1 , x 2 , x 3 ϕ 0 , x 2 , x 3 ϕ x 1 , 1 , x 3 + ϕ x 1 , 0 , x 3 + ϕ x 1 , x 2 , 1 ϕ x 1 , x 2 , 0 . {:[del sigma=(phi(1,x_(2),x_(3))-phi(0,x_(2),x_(3)))-(phi(x_(1),1,x_(3))-phi(x_(1),0,x_(3)))],[+(phi(x_(1),x_(2),1)-phi(x_(1),x_(2),0))],[=phi(1,x_(2),x_(3))-phi(0,x_(2),x_(3))-phi(x_(1),1,x_(3))+phi(x_(1),0,x_(3))],[+phi(x_(1),x_(2),1)-phi(x_(1),x_(2),0).]:}\begin{aligned} \partial \sigma= & \left(\phi\left(1, x_{2}, x_{3}\right)-\phi\left(0, x_{2}, x_{3}\right)\right)-\left(\phi\left(x_{1}, 1, x_{3}\right)-\phi\left(x_{1}, 0, x_{3}\right)\right) \\ & +\left(\phi\left(x_{1}, x_{2}, 1\right)-\phi\left(x_{1}, x_{2}, 0\right)\right) \\ = & \phi\left(1, x_{2}, x_{3}\right)-\phi\left(0, x_{2}, x_{3}\right)-\phi\left(x_{1}, 1, x_{3}\right)+\phi\left(x_{1}, 0, x_{3}\right) \\ & +\phi\left(x_{1}, x_{2}, 1\right)-\phi\left(x_{1}, x_{2}, 0\right) . \end{aligned}σ=(ϕ(1,x2,x3)ϕ(0,x2,x3))(ϕ(x1,1,x3)ϕ(x1,0,x3))+(ϕ(x1,x2,1)ϕ(x1,x2,0))=ϕ(1,x2,x3)ϕ(0,x2,x3)ϕ(x1,1,x3)+ϕ(x1,0,x3)+ϕ(x1,x2,1)ϕ(x1,x2,0).
An example will hopefully clear up the confusion this all was sure to generate:
Fig. 7.3. Orienting the boundary of a 2-cell.
Example 32. Suppose φ ( r , θ ) = ( r cos Γ θ , r sin n θ ) φ ( r , θ ) = ( r cos Γ θ , r sin n θ ) varphi(r,theta)=(r cos Gamma theta,r sin n theta)\varphi(r, \theta)=(r \cos \Gamma \theta, r \sin n \theta)φ(r,θ)=(rcosΓθ,rsinnθ). The image of φ φ varphi\varphiφ is the 2-cell, σ σ sigma\sigmaσ, depicted in Figure 7.3. By the above definition,
σ = ( ϕ ( 1 , θ ) ϕ ( 0 , θ ) ) ( ϕ ( r , 1 ) ϕ ( r , 0 ) ) = ( cos π θ , sin π θ ) ( 0 , 0 ) + ( r , 0 ) ( r , 0 ) . σ = ( ϕ ( 1 , θ ) ϕ ( 0 , θ ) ) ( ϕ ( r , 1 ) ϕ ( r , 0 ) ) = ( cos π θ , sin π θ ) ( 0 , 0 ) + ( r , 0 ) ( r , 0 ) . {:[del sigma=(phi(1","theta)-phi(0","theta))-(phi(r","1)-phi(r","0))],[=(cos pi theta","sin pi theta)-(0","0)+(r","0)-(-r","0).]:}\begin{aligned} \partial \sigma & =(\phi(1, \theta)-\phi(0, \theta))-(\phi(r, 1)-\phi(r, 0)) \\ & =(\cos \pi \theta, \sin \pi \theta)-(0,0)+(r, 0)-(-r, 0) . \end{aligned}σ=(ϕ(1,θ)ϕ(0,θ))(ϕ(r,1)ϕ(r,0))=(cosπθ,sinπθ)(0,0)+(r,0)(r,0).
This is the 1-chain depicted in Figure 7.3.
Finally, we are ready to define what we mean by the boundary of an n n nnn-chain. If C = Σ n i σ j C = Σ n i σ j C=Sigman_(i)sigma_(j)C=\Sigma n_{i} \sigma_{j}C=Σniσj, then we define C = Σ n i σ i C = Σ n i σ i del C=Sigman_(i)delsigma_(i)\partial C=\Sigma n_{i} \partial \sigma_{i}C=Σniσi. Example 33. Suppose
ϕ 1 ( r , θ ) = ( r cos 2 π θ , r sin 2 π θ , 1 r 2 ) , φ 2 ( r , θ ) = ( r cos 2 π θ , r sin 2 π θ , 1 r 2 ) , σ 1 = Im ( φ 1 ) and σ 2 = Im ( φ 2 ) . Then σ 1 + σ 2 is a sphere in R 3 . One can check that ( σ 1 + σ 2 ) = ϕ 1 ( r , θ ) = r cos 2 π θ , r sin 2 π θ , 1 r 2 , φ 2 ( r , θ ) = r cos 2 π θ , r sin 2 π θ , 1 r 2 , σ 1 = Im φ 1  and  σ 2 = Im φ 2 . Then  σ 1 + σ 2  is a sphere in  R 3 .  One   can check that  σ 1 + σ 2 = {:[phi_(1)(r","theta)=(r cos 2pi theta,r sin 2pi theta,sqrt(1-r^(2)))","],[varphi_(2)(r","theta)=(-r cos 2pi theta,r sin 2pi theta,-sqrt()1-r^(2))","],[sigma_(1)=Im(varphi_(1))" and "sigma_(2)=Im(varphi_(2))". Then "sigma_(1)+sigma_(2)" is a sphere in "R^(3)." One "],[" can check that "del(sigma_(1)+sigma_(2))=]:}\begin{aligned} & \phi_{1}(r, \theta)=\left(r \cos 2 \pi \theta, r \sin 2 \pi \theta, \sqrt{1-r^{2}}\right), \\ & \varphi_{2}(r, \theta)=\left(-r \cos 2 \pi \theta, r \sin 2 \pi \theta,-\sqrt{ } 1-r^{2}\right), \\ & \sigma_{1}=\operatorname{Im}\left(\varphi_{1}\right) \text { and } \sigma_{2}=\operatorname{Im}\left(\varphi_{2}\right) \text {. Then } \sigma_{1}+\sigma_{2} \text { is a sphere in } \mathrm{R}^{3} . \text { One } \\ & \text { can check that } \partial\left(\sigma_{1}+\sigma_{2}\right)= \end{aligned}ϕ1(r,θ)=(rcos2πθ,rsin2πθ,1r2),φ2(r,θ)=(rcos2πθ,rsin2πθ,1r2),σ1=Im(φ1) and σ2=Im(φ2). Then σ1+σ2 is a sphere in R3. One  can check that (σ1+σ2)=
7.2. If σ σ sigma\sigmaσ is an n n nnn-cell, show that σ = σ = del del sigma=O/\partial \partial \sigma=\varnothingσ=. (At least show this if σ σ sigma\sigmaσ is a 2 -cell and a 3 -cell. The 2 -cell case can be deduced pictorially from Figures 7.1 and 7.2.)
7.3. If σ σ sigma\sigmaσ is given by the parameterization
ϕ ( r , θ ) = ( r cos θ , r sin θ ) ϕ ( r , θ ) = ( r cos θ , r sin θ ) phi(r,theta)=(r cos theta,r sin theta)\phi(r, \theta)=(r \cos \theta, r \sin \theta)ϕ(r,θ)=(rcosθ,rsinθ)
for 0 r 1 0 r 1 0 <= r <= 10 \leq r \leq 10r1 and 0 θ 4 0 θ 4 0 <= theta <= 40 \leq \theta \leq 40θ4, then what is σ σ del sigma\partial \sigmaσ ?
7.4. If σ σ sigma\sigmaσ is given by the parameterization
ϕ ( r , θ ) = ( r cos θ , r sin θ , r ) ϕ ( r , θ ) = ( r cos θ , r sin θ , r ) phi(r,theta)=(r cos theta,r sin theta,r)\phi(r, \theta)=(r \cos \theta, r \sin \theta, r)ϕ(r,θ)=(rcosθ,rsinθ,r)
for 0 r 1 0 r 1 0 <= r <= 10 \leq r \leq 10r1 and 0 θ 2 π 0 θ 2 π 0 <= theta <= 2pi0 \leq \theta \leq 2 \pi0θ2π, then what is σ σ del sigma\partial \sigmaσ ?

7.2 The generalized Stokes' Theorem

In calculus, we learn that when you take a function, differentiate it, and then integrate the result, something special happens. In this section, we explore what happens when we take a form, differentiate it, and then integrate the resulting form over some chain. The general argument is quite complicated, so we start by looking at forms of a particular type integrated over very special regions.
Suppose ω = a d x 2 d x 3 ω = a d x 2 d x 3 omega=adx_(2)^^dx_(3)\omega=a d x_{2} \wedge d x_{3}ω=adx2dx3 is a 2-form on R 3 R 3 R^(3)R^{3}R3, where a : R 3 R a : R 3 R a:R^(3)rarr Ra: R^{3} \rightarrow Ra:R3R. Let R R RRR be the unit cube, P 3 R 3 P 3 R 3 P^(3)subR^(3)P^{3} \subset R^{3}P3R3. We would like to explore what happens when we integrate ω ω ω ω omega omega\omega \omegaωω over R R RRR. Note first that Problem 6.8 implies that d ω = a x 1 d x 1 d x 2 d x 3 d ω = a x 1 d x 1 d x 2 d x 3 ^(d omega)=(del a)/(delx_(1))dx_(1)^^dx_(2)^^dx_(3){ }^{d \omega}=\frac{\partial a}{\partial x_{1}} d x_{1} \wedge d x_{2} \wedge d x_{3}dω=ax1dx1dx2dx3.
Recall the steps used to define R a w R a w int Raw\int R a wRaw :
  1. . Choose a lattice of points in R , { p i , j , k } R , p i , j , k R,{p_(i,j,k)}R,\left\{p_{i, j, k}\right\}R,{pi,j,k}. Since R R RRR is a cube, we can choose this lattice to be rectangular.
  2. Define V 1 i , j , k = p i + 1 , j , k p i , j , k V 1 i , j , k = p i + 1 , j , k p i , j , k V^(1)_(i,j,k)=p_(i+1,j,k)-p_(i,j,k)V^{1}{ }_{i, j, k}=p_{i+1, j, k}-p_{i, j, k}V1i,j,k=pi+1,j,kpi,j,k. Similarly, define V i , j , k 2 V i , j , k 2 V_(i,j,k)^(2)V_{i, j, k}^{2}Vi,j,k2 and V i , j , k V i , j , k V_(i,j,k)V_{i, j, k}Vi,j,k
  3. Compute a ω p i , j , k ( V i , j , k V i , j , k V i , j , k 2 ) a ω p i , j , k V i , j , k V i , j , k V i , j , k 2 a omegap_(i,j,k)(V_(i,j,k)V_(i,j,k)V_(i,j,k)^(2))a \omega p_{i, j, k}\left(V_{i, j, k} V_{i, j, k} V_{i, j, k}^{2}\right)aωpi,j,k(Vi,j,kVi,j,kVi,j,k2).
  4. Sum over all i , j i , j i,ji, ji,j and k k kkk.
  5. Take the limit as the maximal distance between adjacent lattice points goes to zero.
Let's focus on Step 3 for a moment. Let t t ttt be the distance between p i + 1 , j , k p i + 1 , j , k p_(i+1,j,k)p_{i+1, j, k}pi+1,j,k and p i , j , k p i , j , k p_(i,j,k')p_{i, j, k \prime}pi,j,k and assume t t ttt is small. Then a x 1 ( p i , j , k ) is a x 1 p i , j , k is  (del a)/(delx_(1))(p_(i,j,k))_("is ")\frac{\partial a}{\partial x_{1}}\left(p_{i, j, k}\right)_{\text {is }}ax1(pi,j,k)is  approximately equal to a ( p i + 1 , j , k ) a ( p i , j , k ) t a p i + 1 , j , k a p i , j , k t (a(p_(i+1,j,k))-a(p_(i,j,k)))/(t)\frac{a\left(p_{i+1, j, k}\right)-a\left(p_{i, j, k}\right)}{t}a(pi+1,j,k)a(pi,j,k)t. This approximation gets better and better when we let t 0 t 0 t rarr0t \rightarrow 0t0, in Step 5 .
The vectors, V 1 i , j , k V 1 i , j , k V^(1)_(i,j,k)V^{1}{ }_{i, j, k}V1i,j,k through V i , j , k 3 V i , j , k 3 V_(i,j,k)^(3)V_{i, j, k}{ }^{3}Vi,j,k3, form a little cube. If we say the vector V i , j , k 1 V i , j , k 1 V_(i,j,k)^(1)V_{i, j, k}{ }^{1}Vi,j,k1 is "vertical," and the other two are horizontal, then the "height" of this cube is t t ttt, and the area of its base is d x 2 d x 3 ( V i , j , k 2 d x 2 d x 3 V i , j , k 2 dx_(2)^^dx_(3)(V_(i,j,k)^(2):}d x_{2} \wedge d x_{3}\left(V_{i, j, k}{ }^{2}\right.dx2dx3(Vi,j,k2 V 3 i , j , k ) V 3 i , j , k {:V^(3)_(i,j,k))\left.V^{3}{ }_{i, j, k}\right)V3i,j,k), which makes its volume t d x 2 d x 3 ( V i , j , k 2 V i , j , k 3 ) t d x 2 d x 3 V i , j , k 2 V i , j , k 3 tdx_(2)^^dx_(3)(V_(i,j,k)^(2)V_(i,j,k)^(3))t d x_{2} \wedge d x_{3}\left(V_{i, j, k}^{2} V_{i, j, k}^{3}\right)tdx2dx3(Vi,j,k2Vi,j,k3). Putting all this together, we find that
d ω p i , j , k ( V i , j , k 1 , V i , j , k 2 , V i , j , k 2 ) = a x 1 d x 1 d x 2 d x 3 ( V i , j , k 1 , V i , j , k 2 , V i , j , k 2 ) a ( p i + 1 , j , k ) a ( p i , j , k ) t t d x 2 d x 3 ( V i , j , k 2 , V i , j , k 3 ) = ω ( V i + 1 , j , k 2 , V i + 1 , j , k 3 ) ω ( V i , j , k 2 , V i , j , k 3 ) d ω p i , j , k V i , j , k 1 , V i , j , k 2 , V i , j , k 2 = a x 1 d x 1 d x 2 d x 3 V i , j , k 1 , V i , j , k 2 , V i , j , k 2 a p i + 1 , j , k a p i , j , k t t d x 2 d x 3 V i , j , k 2 , V i , j , k 3 = ω V i + 1 , j , k 2 , V i + 1 , j , k 3 ω V i , j , k 2 , V i , j , k 3 {:[domega_(p_(i,j,k))(V_(i,j,k)^(1),V_(i,j,k)^(2),V_(i,j,k)^(2))=(del a)/(delx_(1))dx_(1)^^dx_(2)^^dx_(3)(V_(i,j,k)^(1),V_(i,j,k)^(2),V_(i,j,k)^(2))],[~~(a(p_(i+1,j,k))-a(p_(i,j,k)))/(t)tdx_(2)^^dx_(3)(V_(i,j,k)^(2),V_(i,j,k)^(3))],[=omega(V_(i+1,j,k)^(2),V_(i+1,j,k)^(3))-omega(V_(i,j,k)^(2),V_(i,j,k)^(3))]:}\begin{aligned} d \omega_{p_{i, j, k}}\left(V_{i, j, k}^{1}, V_{i, j, k}^{2}, V_{i, j, k}^{2}\right) & =\frac{\partial a}{\partial x_{1}} d x_{1} \wedge d x_{2} \wedge d x_{3}\left(V_{i, j, k}^{1}, V_{i, j, k}^{2}, V_{i, j, k}^{2}\right) \\ & \approx \frac{a\left(p_{i+1, j, k}\right)-a\left(p_{i, j, k}\right)}{t} t d x_{2} \wedge d x_{3}\left(V_{i, j, k}^{2}, V_{i, j, k}^{3}\right) \\ & =\omega\left(V_{i+1, j, k}^{2}, V_{i+1, j, k}^{3}\right)-\omega\left(V_{i, j, k}^{2}, V_{i, j, k}^{3}\right) \end{aligned}dωpi,j,k(Vi,j,k1,Vi,j,k2,Vi,j,k2)=ax1dx1dx2dx3(Vi,j,k1,Vi,j,k2,Vi,j,k2)a(pi+1,j,k)a(pi,j,k)ttdx2dx3(Vi,j,k2,Vi,j,k3)=ω(Vi+1,j,k2,Vi+1,j,k3)ω(Vi,j,k2,Vi,j,k3)
Let's move on to Step 4. Here we sum over all i , j i , j i,ji, ji,j and k k kkk. Suppose for the moment that i i iii ranges between 1 and N N NNN. First, we fix j j jjj and k k kkk, and sum over all i i iii. The result is ω ( V N , j , k 2 V N , j , k ) ω ( V 1 , j , k 2 V 1 , j , k ) ω V N , j , k 2 V N , j , k ω V 1 , j , k 2 V 1 , j , k omega(V_(N,j,k)^(2)V_(N,j,k))-omega(V_(1,j,k)^(2)V_(1,j,k))\omega\left(V_{N, j, k}^{2} V_{N, j, k}\right)-\omega\left(V_{1, j, k}^{2} V_{1, j, k}\right)ω(VN,j,k2VN,j,k)ω(V1,j,k2V1,j,k). Now notice that j , k ω ( V 2 N , j , k , V N , j , k 3 ) j , k ω V 2 N , j , k , V N , j , k 3 sumj,komega(V^(2)_(N,j,k),V_(N,j,k)^(3))\sum \mathrm{j}, \mathrm{k} \omega\left(V^{2}{ }_{N, j, k}, V_{N, j, k}^{3}\right)j,kω(V2N,j,k,VN,j,k3) is a Riemann sum for the integral of ω ω omega\omegaω over the "top" of R R RRR, and j , k ω ( V 1 , j , k 2 , V 1 , j , k 3 ) j , k ω V 1 , j , k 2 , V 1 , j , k 3 sum_(j,k)omega(V_(1,j,k)^(2),V_(1,j,k)^(3))\sum_{j, k} \omega\left(V_{1, j, k}^{2}, V_{1, j, k}^{3}\right)j,kω(V1,j,k2,V1,j,k3) is a Riemann sum for ω ω omega\omegaω over the "bottom" of R R RRR. Lastly, not that ω ω omega\omegaω, evaluated on any pair of vectors which lie in the sides of the cube, gives zero. Hence, the integral of ω ω omega\omegaω over a side of R R RRR is zero. Putting all this together, we conclude:
(7.1) R d ω = R ω . (7.1) R d ω = R ω . {:(7.1)int_(R)d omega=int_(del R)omega.:}\begin{equation*} \int_{R} d \omega=\int_{\partial R} \omega . \tag{7.1} \end{equation*}(7.1)Rdω=Rω.
7.5. Prove that Equation 7.1 holds if ω = b d x 1 d x 3 ω = b d x 1 d x 3 omega=bdx_(1)^^dx_(3)\omega=b d x_{1} \wedge d x_{3}ω=bdx1dx3, or if ω = c ω = c omega=c\omega=cω=c d x 1 d x 2 d x 1 d x 2 dx_(1)^^dx_(2)d x_{1} \wedge d x_{2}dx1dx2. Caution! Beware of signs and orientations.
7.6. Use the previous problem to conclude that if ω = a d x 2 d x 3 + ω = a d x 2 d x 3 + omega=adx_(2)^^dx_(3)+\omega=a d x_{2} \wedge d x_{3}+ω=adx2dx3+ b d x 1 d x 3 + c d x 1 d x 2 b d x 1 d x 3 + c d x 1 d x 2 bdx_(1)^^dx_(3)+cdx_(1)^^dx_(2)b d x_{1} \wedge d x_{3}+c d x_{1} \wedge d x_{2}bdx1dx3+cdx1dx2 is an arbitrary 2-form on R 3 R 3 R^(3)\mathrm{R}^{3}R3, then Equation 7.1 holds.
7.7. If ω ω omega\omegaω is an arbitrary ( n 1 n 1 n-1n-1n1 )-form on R n R n R^(n)\mathrm{R}^{n}Rn and R R RRR is the unit cube in R n R n R^(n)\mathrm{R}^{n}Rn, then show that Equation 7.1 still holds.
In general, if C = n σ i C = n σ i C=sum nsigma_(i)C=\sum n \sigma_{i}C=nσi is an n n nnn-chain, then
C ω = C d ω . C ω = C d ω int_(del C)omega=int_(C)d omega". "\int_{\partial C} \omega=\int_{C} d \omega \text {. }Cω=Cdω
This equation is called the generalized Stokes' Theorem. It is unquestionably the most crucial result of this text. In fact, everything we have done up to this point has been geared toward developing this equation and everything that follows will be applications of this equation. Technically, we have only established this theorem when integrating over cubes and their boundaries. We postpone the general proof to Section 9.1.
Example 34. Let ω = x d y ω = x d y omega=xdy\omega=x d yω=xdy be a 1 -form on R 2 R 2 R^(2)\mathrm{R}^{2}R2. Let σ σ sigma\sigmaσ be the 2 -cell which is the image of the parameterization φ ( r , θ ) = ( r cos θ , r sin θ ) φ ( r , θ ) = ( r cos θ , r sin θ ) varphi(r,theta)=(r cos theta,r sin theta)\varphi(r, \theta)=(r \cos \theta, r \sin \theta)φ(r,θ)=(rcosθ,rsinθ), where 0 r R 0 r R 0 <= r <= R0 \leq r \leq R0rR and 0 θ 2 π 0 θ 2 π 0 <= theta <= 2pi0 \leq \theta \leq 2 \pi0θ2π. By the generalized Stokes' Theorem,
σ ω = σ d ω = σ d x d y = σ d x d y = Area ( σ ) = π R 2 . σ ω = σ d ω = σ d x d y = σ d x d y = Area ( σ ) = π R 2 . int_(del sigma)omega=int_(sigma)d omega=int_(sigma)dx^^dy=int_(sigma)dxdy=Area(sigma)=piR^(2).\int_{\partial \sigma} \omega=\int_{\sigma} d \omega=\int_{\sigma} d x \wedge d y=\int_{\sigma} d x d y=\operatorname{Area}(\sigma)=\pi R^{2} .σω=σdω=σdxdy=σdxdy=Area(σ)=πR2.
7.8. Verify directly that σ ω = π R 2 σ ω = π R 2 int del sigma omega=piR^(2)\int \partial \sigma \omega=\pi R^{2}σω=πR2.
Example 35. Let ω = x d y + y d x ω = x d y + y d x omega=xdy+ydx\omega=x d y+y d xω=xdy+ydx be a 1-form on R 2 R 2 R^(2)\mathrm{R}^{2}R2, and let σ σ sigma\sigmaσ be any 2 -cell. Then σ ω = σ a ω = 0 σ ω = σ a ω = 0 int del sigma omega=int sigma a omega=0\int \partial \sigma \omega=\int \sigma a \omega=0σω=σaω=0.
7.9. Find a 1-chain in R 2 R 2 R^(2)R^{2}R2, which bounds a 2-cell, and integrate the form x d y + y d x x d y + y d x xdy+ydxx d y+y d xxdy+ydx over this curve.
7.10. Let ω ω omega\omegaω be a differential ( n 1 n 1 n-1n-1n1 )-form and σ σ sigma\sigmaσ a ( n + 1 n + 1 n+1n+1n+1 )-cell. Use the generalized Stokes' Theorem in two different ways to show d d int d\int dd ω = 0 . σ ^ σ ω = 0 . σ ^ σ omega=0. hat(sigma)sigma\omega=0 . \hat{\sigma} \sigmaω=0.σ^σ
Example 36. Let C C CCC be the curve in R 2 R 2 R^(2)\mathrm{R}^{2}R2 parameterized by φ ( t ) = ( t 2 φ ( t ) = t 2 varphi(t)=(t^(2):}\varphi(t)=\left(t^{2}\right.φ(t)=(t2, t 3 t 3 t^(3)t^{3}t3 ), where 1 t 1 1 t 1 -1 <= t <= 1-1 \leq t \leq 11t1. Let f f fff be the 0 -form x 2 y x 2 y x^(2)yx^{2} yx2y. We use the generalized Stokes' Theorem to calculate SCdf.
The curve C C CCC goes from the point ( 1 , 1 ) ( 1 , 1 ) (1,-1)(1,-1)(1,1), when t = 1 t = 1 t=-1t=-1t=1, to the point ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1), when t = 1 t = 1 t=1t=1t=1. Hence, C C del C\partial CC is the 0 -chain ( 1 , 1 ) ( 1 , 1 ) ( 1 , 1 ) ( 1 , 1 ) (1,1)-(1,-1)(1,1)-(1,-1)(1,1)(1,1). Now we use Stokes:
C d f = C f = ( 1 , 1 ) ( 1 , 1 ) x 2 y = 1 ( 1 ) = 2 . C d f = C f = ( 1 , 1 ) ( 1 , 1 ) x 2 y = 1 ( 1 ) = 2 . int_(C)df=int_(del C)f=int_((1,1)-(1,-1))x^(2)y=1-(-1)=2.\int_{C} d f=\int_{\partial C} f=\int_{(1,1)-(1,-1)} x^{2} y=1-(-1)=2 .Cdf=Cf=(1,1)(1,1)x2y=1(1)=2.
7.11. Calculate C d f C d f intCdf\int \mathrm{C} d fCdf directly.
7.12. Let C C CCC be any curve in R 3 R 3 R^(3)\mathrm{R}^{3}R3 from ( 0 , 0 , 0 ) ( 0 , 0 , 0 ) (0,0,0)(0,0,0)(0,0,0) to ( 1 , 1 , 1 ) ( 1 , 1 , 1 ) (1,1,1)(1,1,1)(1,1,1). Let ω = ω = omega=\omega=ω= y 2 z 2 d x + 2 x y z 2 d y + 2 x y 2 z d z y 2 z 2 d x + 2 x y z 2 d y + 2 x y 2 z d z y^(2)z^(2)dx+2xyz^(2)dy+2xy^(2)zdzy^{2} z^{2} d x+2 x y z^{2} d y+2 x y^{2} z d zy2z2dx+2xyz2dy+2xy2zdz. Calculate C ω C ω intComega\int \mathrm{C} \omegaCω.
Example 37. Let ω = ( x 2 + y ) d x + ( x y 2 ) d y ω = x 2 + y d x + x y 2 d y omega=(x^(2)+y)dx+(x-y^(2))dy\omega=\left(x^{2}+y\right) d x+\left(x-y^{2}\right) d yω=(x2+y)dx+(xy2)dy be a 1 -form on R 2 R 2 R^(2)\mathrm{R}^{2}R2. We wish to integrate ω ω omega\omegaω over σ σ sigma\sigmaσ, the top half of the unit circle, oriented clockwise. First, note that a ω = 0 a ω = 0 a omega=0a \omega=0aω=0, so that if we integrate ω ω omega\omegaω over the boundary of any 2 -cell, we would get zero. Let T denote the line segment connecting ( 1 , 0 ) ( 1 , 0 ) (-1,0)(-1,0)(1,0) to ( 1,0 ). Then the 1 -chain σ T σ T sigma-T\sigma-TσT bounds a 2 -cell. So σ T ω = 0 σ T ω = 0 int sigma-T omega=0\int \sigma-T \omega=0σTω=0, which implies that σ ω = T ω σ ω = T ω int sigma omega=int T omega\int \sigma \omega=\int T \omegaσω=Tω. This latter integral is a bit easier to com ute. Let φ ( t ) = ( t , 0 ) φ ( t ) = ( t , 0 ) varphi(t)=(t,0)\varphi(t)=(t, 0)φ(t)=(t,0) be a parame erization of T , where 1 t 1 1 t 1 -1 <= t <= 1-1 \leq t \leq 11t1. Then
σ ω = τ ω = [ 1 , 1 ] ω ( t , 0 ) ( 1 , 0 ) d t = 1 1 t 2 d t = 2 3 σ ω = τ ω = [ 1 , 1 ] ω ( t , 0 ) ( 1 , 0 ) d t = 1 1 t 2 d t = 2 3 int_(sigma)omega=int_(tau)omega=int_([-1,1])omega_((t,0))((:1,0:))dt=int_(-1)^(1)t^(2)dt=(2)/(3)\int_{\sigma} \omega=\int_{\tau} \omega=\int_{[-1,1]} \omega_{(t, 0)}(\langle 1,0\rangle) d t=\int_{-1}^{1} t^{2} d t=\frac{2}{3}σω=τω=[1,1]ω(t,0)(1,0)dt=11t2dt=23
7.13. Let ω = y 2 d x + x 2 d y ω = y 2 d x + x 2 d y omega=-y^(2)dx+x^(2)dy\omega=-y^{2} d x+x^{2} d yω=y2dx+x2dy. Let σ σ sigma\sigmaσ be the 2 -cell in R 2 R 2 R^(2)\mathrm{R}^{2}R2 parameterized by the following:
ϕ ( u , v ) = ( 2 u v , u + v ) , 1 u 2 , 0 v 1 . ϕ ( u , v ) = ( 2 u v , u + v ) , 1 u 2 , 0 v 1 . phi(u,v)=(2u-v,u+v),1 <= u <= 2,0 <= v <= 1.\phi(u, v)=(2 u-v, u+v), 1 \leq u \leq 2,0 \leq v \leq 1 .ϕ(u,v)=(2uv,u+v),1u2,0v1.
Calculate ω . δ σ ω . δ σ int omega.delta sigma\int \omega . \delta \sigmaω.δσ
7.14. Let ω = d x ln x d y ω = d x ln x d y omega=dx-ln xdy\omega=d x-\ln x d yω=dxlnxdy. Let σ σ sigma\sigmaσ be the 2-cell parameterized by the following:
ϕ ( u , v ) = ( u v 2 , u 3 v ) , 1 u 2 , 1 v 2 . ϕ ( u , v ) = u v 2 , u 3 v , 1 u 2 , 1 v 2 . phi(u,v)=(uv^(2),u^(3)v),1 <= u <= 2,1 <= v <= 2.\phi(u, v)=\left(u v^{2}, u^{3} v\right), 1 \leq u \leq 2,1 \leq v \leq 2 .ϕ(u,v)=(uv2,u3v),1u2,1v2.
Calculate: ω ω int omega\int \omegaω. σ ¯ σ σ ¯ σ bar(sigma)sigma\bar{\sigma} \sigmaσ¯σ
7.15. Let σ σ sigma\sigmaσ be the 2-cell given by the following parameterization:
ϕ ( r , θ ) = ( r cos θ , r sin θ ) , 0 r 1 , 0 θ π . ϕ ( r , θ ) = ( r cos θ , r sin θ ) , 0 r 1 , 0 θ π . phi(r,theta)=(r cos theta,r sin theta),0 <= r <= 1,0 <= theta <= pi.\phi(r, \theta)=(r \cos \theta, r \sin \theta), 0 \leq r \leq 1,0 \leq \theta \leq \pi .ϕ(r,θ)=(rcosθ,rsinθ),0r1,0θπ.
Suppose ω = x 2 d x + e y d y ω = x 2 d x + e y d y omega=x^(2)dx+e^(y)dy\omega=x^{2} d x+e^{y} d yω=x2dx+eydy.
  1. Calculate σ a ω σ a ω int sigma a omega\int \sigma a \omegaσaω directly.
  2. Let C 1 C 1 C_(1)C_{1}C1 be the horizontal segment connecting ( 1 , 0 ) ( 1 , 0 ) (-1,0)(-1,0)(1,0) to ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0), and C 2 C 2 C_(2)C_{2}C2 be the horizontal segment connecting ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0) to ( 1 , 0 ) ( 1 , 0 ) (1,0)(1,0)(1,0). Calculate C 1 ω C 1 ω intC_(1)omega\int \mathrm{C}_{1} \omegaC1ω and C 1 ω C 1 ω intC_(1)omega\int \mathrm{C}_{1} \omegaC1ω directly.
  3. Use your previous answers to determine the integral of ω ω omega\omegaω over the top half of the unit circle (oriented counterclockwise).
    7.16. Let ω = ( x + y 3 ) d x + 3 x y 2 d y ω = x + y 3 d x + 3 x y 2 d y omega=(x+y^(3))dx+3xy^(2)dy\omega=\left(x+y^{3}\right) d x+3 x y^{2} d yω=(x+y3)dx+3xy2dy be a differential 1-form on R 2 R 2 R^(2)\mathrm{R}^{2}R2. Let Q Q QQQ be the rectangle { ( x , y ) 0 x 3 , 0 y 2 } { ( x , y ) 0 x 3 , 0 y 2 } {(x,y)∣0 <= x <= 3,0 <= y <= 2}\{(x, y) \mid 0 \leq x \leq 3,0 \leq y \leq 2\}{(x,y)0x3,0y2}.
  4. Compute a b a b aba bab.
  5. Use the generalized Stokes' Theorem to compute del\partial.
  6. Compute del\partial directly, by integrating ω ω omega\omegaω over each each edge of the boundary Q Q del Q\partial QQ of the rect ngle, and then adding in the appropriate manner.
  7. How does T L ω T L ω int_(-T-L)omega\int_{-T-L} \omegaTLω compare to B ω B ω int_(B)omega\int_{B} \omegaBω ?
  8. Let S S SSS be any curve in the upper half plane (i.e., the set { ( x { ( x {(x\{(x{(x, y ) y 0 } ) y ) y 0 } ) y)∣y >= 0})y) \mid y \geq 0\})y)y0}) that goes from the point ( 3 , 0 ) ( 3 , 0 ) (3,0)(3,0)(3,0) to the point ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0). What iss ? Why?
  9. Let S S SSS be any curve that goes from the point ( 3 , 0 ) ( 3 , 0 ) (3,0)(3,0)(3,0) to the point ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0). What iss ω ω ^(omega){ }^{\omega}ω ? WHY?
    7.17. Let ω ω omega\omegaω be the following 2 -form on R 3 R 3 R^(3)\mathbb{R}^{3}R3 :
ω = ( x 2 + y 2 ) d y d z + ( x 2 y 2 ) d x d z . ω = x 2 + y 2 d y d z + x 2 y 2 d x d z . omega=(x^(2)+y^(2))dy^^dz+(x^(2)-y^(2))dx^^dz.\omega=\left(x^{2}+y^{2}\right) d y \wedge d z+\left(x^{2}-y^{2}\right) d x \wedge d z .ω=(x2+y2)dydz+(x2y2)dxdz.
Let V V VVV be the region of R 3 R 3 R^(3)\mathrm{R}^{3}R3 bounded by the graph of y = 1 x 2 y = 1 x 2 y=sqrt()1-x^(2)y=\sqrt{ } 1-x^{2}y=1x2, the planes z = 0 z = 0 z=0z=0z=0 and z = 2 z = 2 z=2z=2z=2, and the x z x z xzx zxz-plane (see Figure 7.4).
Fig. 7.4. The region V V VVV of Problem 7.17.
  1. Parameterize V V VVV using cylindrical coordinates.
  2. Determine a ω a ω a omegaa \omegaaω.
  3. Calculatev.
  4. The sides f V f V fV\mathrm{f} VfV are parameterized as follows:
    a. Bottom: φ B ( r , θ ) = ( r cos θ , r sin θ , 0 ) φ B ( r , θ ) = ( r cos θ , r sin θ , 0 ) varphi_(B)(r,theta)=(r cos theta,r sin theta,0)\varphi_{B}(r, \theta)=(r \cos \theta, r \sin \theta, 0)φB(r,θ)=(rcosθ,rsinθ,0), where 0 r 0 r 0 <= r0 \leq r0r 1 1 <= 1\leq 11 and 0 θ π 0 θ π 0 <= theta <= pi0 \leq \theta \leq \pi0θπ.
    b. Top: φ T ( r , θ ) = ( r cos θ , r sin θ , 2 ) φ T ( r , θ ) = ( r cos θ , r sin θ , 2 ) varphi_(T)(r,theta)=(r cos theta,r sin theta,2)\varphi_{T}(r, \theta)=(r \cos \theta, r \sin \theta, 2)φT(r,θ)=(rcosθ,rsinθ,2), where 0 r 0 r 0 <= r <=0 \leq r \leq0r 1 and 0 θ π 0 θ π 0 <= theta <= pi0 \leq \theta \leq \pi0θπ.
    c. Flat side: φ F ( x , z ) = ( x , 0 , z ) φ F ( x , z ) = ( x , 0 , z ) varphi_(F)(x,z)=(x,0,z)\varphi_{F}(x, z)=(x, 0, z)φF(x,z)=(x,0,z), where 1 x 1 1 x 1 -1 <= x <= 1-1 \leq x \leq 11x1 and 0 z 2 z 2 <= z <= 2\leq z \leq 2z2.
    d. Curved side: φ C ( θ , z ) = ( cos θ , sin θ , z ) φ C ( θ , z ) = ( cos θ , sin θ , z ) varphi_(C)(theta,z)=(cos theta,sin theta,z)\varphi_{C}(\theta, z)=(\cos \theta, \sin \theta, z)φC(θ,z)=(cosθ,sinθ,z), where 0 0 0 <=0 \leq0 θ π θ π theta <= pi\theta \leq \piθπ and 0 z 2 0 z 2 0 <= z <= 20 \leq z \leq 20z2.
Calculate the integral of ω ω omega\omegaω over the top, bottom and flat side. (Do not calculate this integral over the curved side.)
5. If C C CCC is the curved side of V V del V\partial VV, use your answers to the previous questions to determine C C CCC
7.18. Calculate the volume of a ball of radius one, { ( ρ , θ , φ ) ρ { ( ρ , θ , φ ) ρ {(rho,theta,varphi)∣rho <=\{(\rho, \theta, \varphi) \mid \rho \leq{(ρ,θ,φ)ρ 1 } 1 } 1}1\}1}, by integrating some 2 -form over the sphere of radius one, { ( ρ { ( ρ {(rho\{(\rho{(ρ, θ , φ ) ρ = 1 } θ , φ ) ρ = 1 } theta,varphi)∣rho=1}\theta, \varphi) \mid \rho=1\}θ,φ)ρ=1}.
7.19. Calculate where C C CCC is the circle of radius two, centered about the origin.
C x 3 d x + ( 1 3 x 3 + x y 2 ) d y C x 3 d x + 1 3 x 3 + x y 2 d y int_(C)x^(3)dx+((1)/(3)x^(3)+xy^(2))dy\int_{C} x^{3} d x+\left(\frac{1}{3} x^{3}+x y^{2}\right) d yCx3dx+(13x3+xy2)dy
7.20. Suppose ω = x d x + x d y ω = x d x + x d y omega=xdx+xdy\omega=x d x+x d yω=xdx+xdy is a 1 -form on R 2 R 2 R^(2)\mathrm{R}^{2}R2. Let C C CCC be the ellipse x 2 4 + y 2 9 = 1 x 2 4 + y 2 9 = 1 (x^(2))/(4)+(y^(2))/(9)=1\frac{x^{2}}{4}+\frac{y^{2}}{9}=1x24+y29=1. Determine the value of C C CCC by integrating some 2 -form over the region bounded by the ellipse.
7.21. Let ω = y 2 d x + x 2 d y ω = y 2 d x + x 2 d y omega=-y^(2)dx+x^(2)dy\omega=-y^{2} d x+x^{2} d yω=y2dx+x2dy. Let σ σ sigma\sigmaσ be the 2 -cell in R 2 R 2 R^(2)\mathrm{R}^{2}R2 parameterized by the following:
ϕ ( r , θ ) = ( r cosh θ , r sinh θ ) ϕ ( r , θ ) = ( r cosh θ , r sinh θ ) phi(r,theta)=(r cosh theta,r sinh theta)\phi(r, \theta)=(r \cosh \theta, r \sinh \theta)ϕ(r,θ)=(rcoshθ,rsinhθ)
where 0 r 1 0 r 1 0 <= r <= 10 \leq r \leq 10r1 and 1 θ 1 1 θ 1 -1 <= theta <= 1-1 \leq \theta \leq 11θ1. Calculate ω ω int omega\int \omegaω. σ σ del sigma\partial \sigmaσ
( Recall: cosh θ = e θ + e θ 2 , sinh θ = e θ e θ 2 . )  Recall:  cosh θ = e θ + e θ 2 , sinh θ = e θ e θ 2 . (" Recall: "cosh theta=(e^(theta)+e^(-theta))/(2),quad sinh theta=(e^(theta)-e^(-theta))/(2).)\left(\text { Recall: } \cosh \theta=\frac{e^{\theta}+e^{-\theta}}{2}, \quad \sinh \theta=\frac{e^{\theta}-e^{-\theta}}{2} .\right)( Recall: coshθ=eθ+eθ2,sinhθ=eθeθ2.)
7.22. Suppose ω ω omega\omegaω is a 1 -form on R 2 R 2 R^(2)R^{2}R2 such that a ω = 0 a ω = 0 a omega=0a \omega=0aω=0. Let C 1 C 1 C_(1)C_{1}C1 and C 2 C 2 C_(2)C_{2}C2 be the 1 -cells given by the following parameterizations:
C 1 : ϕ ( t ) = ( t , 0 ) , 2 π t 6 π C 2 : ψ ( t ) = ( t cos t , t sin t ) , 2 π t 6 π C 1 : ϕ ( t ) = ( t , 0 ) , 2 π t 6 π C 2 : ψ ( t ) = ( t cos t , t sin t ) , 2 π t 6 π {:[C_(1):phi(t)=(t","0)","2pi <= t <= 6pi],[C_(2):psi(t)=(t cos t","t sin t)","2pi <= t <= 6pi]:}\begin{aligned} & C_{1}: \phi(t)=(t, 0), 2 \pi \leq t \leq 6 \pi \\ & C_{2}: \psi(t)=(t \cos t, t \sin t), 2 \pi \leq t \leq 6 \pi \end{aligned}C1:ϕ(t)=(t,0),2πt6πC2:ψ(t)=(tcost,tsint),2πt6π
Show that C 1 ω = C 2 ω . C 1 C 2 Show that  C 1 ω = C 2 ω . C 1 C 2 int_("Show that "C_(1))omega=int_(C_(2))omega.C_(1)C_(2)\int_{\text {Show that } C_{1}} \omega=\int_{C_{2}} \omega . C_{1} C_{2}Show that C1ω=C2ω.C1C2
(Cautio : Beware of orientations!)

7.3 Vector calculus and the many faces of the generalized Stokes' Theorem

Although the language and notation may be new, you have already seen the generalized Stokes' Theorem in many guises. For example, let f ( x ) f ( x ) f(x)f(x)f(x) be a 0 -form on
R. Then d f = f ( x ) d x d f = f ( x ) d x df=f^(')(x)dxd f=f^{\prime}(x) d xdf=f(x)dx. Let [ a , b ] [ a , b ] [a,b][a, b][a,b] be a 1 1 1-1-1 cell in R R RRR. Then the generalized Stokes' Theorem tells us which is, of course, the "Fundamental Theorem of Calculus." If we let R R RRR be some 2 -chain in R 2 R 2 R^(2)\mathrm{R}^{2}R2 then the generalized Stokes' Theorem implies
a b f ( x ) d x = [ a , b ] f ( x ) d x = [ a , b ] f ( x ) = b a f ( x ) = f ( b ) f ( a ) , R P d x + Q d y = R d ( P d x + Q d y ) = R ( Q x P y ) d x d y . a b f ( x ) d x = [ a , b ] f ( x ) d x = [ a , b ] f ( x ) = b a f ( x ) = f ( b ) f ( a ) , R P d x + Q d y = R d ( P d x + Q d y ) = R Q x P y d x d y . {:[int_(a)^(b)f^(')(x)dx=int_([a,b])f^(')(x)dx=int_(del[a,b])f(x)=int_(b-a)f(x)=f(b)-f(a)","],[int_(del R)Pdx+Qdy=int_(R)d(Pdx+Qdy)=int_(R)((del Q)/(del x)-(del P)/(del y))dxdy.]:}\begin{aligned} & \int_{a}^{b} f^{\prime}(x) d x=\int_{[a, b]} f^{\prime}(x) d x=\int_{\partial[a, b]} f(x)=\int_{b-a} f(x)=f(b)-f(a), \\ & \int_{\partial R} P d x+Q d y=\int_{R} d(P d x+Q d y)=\int_{R}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y . \end{aligned}abf(x)dx=[a,b]f(x)dx=[a,b]f(x)=baf(x)=f(b)f(a),RPdx+Qdy=Rd(Pdx+Qdy)=R(QxPy)dxdy.
This is what we call "Green's theorem" in calculus. To proceed further, we restrict ourselves to R 3 R 3 R^(3)R^{3}R3. In this dimension, there is a nice correspondence between vector fields and both 1-and 2-forms.
F = F x , F y , F z ω F 1 = F x d x + F y d y + F z d z ω F 2 = F x d y d z F y d x d z + F z d x d y . F = F x , F y , F z ω F 1 = F x d x + F y d y + F z d z ω F 2 = F x d y d z F y d x d z + F z d x d y . {:[F=(:F_(x),F_(y),F_(z):)harromega_(F)^(1)],[=F_(x)dx+F_(y)dy+F_(z)dz],[ harromega_(F)^(2)]:}=F_(x)dy^^dz-F_(y)dx^^dz+F_(z)dx^^dy.\begin{aligned} & \mathbf{F}=\left\langle F_{x}, F_{y}, F_{z}\right\rangle \leftrightarrow \omega_{\mathbf{F}}^{1} \\ &=F_{x} d x+F_{y} d y+F_{z} d z \\ & \leftrightarrow \omega_{\mathbf{F}}^{2} \end{aligned}=F_{x} d y \wedge d z-F_{y} d x \wedge d z+F_{z} d x \wedge d y .F=Fx,Fy,FzωF1=Fxdx+Fydy+FzdzωF2=FxdydzFydxdz+Fzdxdy.
On R 3 R 3 R^(3)R^{3}R3 there is also a useful correspondence between 0-forms (functions) and 3-forms.
f ( x , y , z ) ω f 3 = f d x d y d z . f ( x , y , z ) ω f 3 = f d x d y d z . f(x,y,z)harromega_(f)^(3)=fdx^^dy^^dz.f(x, y, z) \leftrightarrow \omega_{f}^{3}=f d x \wedge d y \wedge d z .f(x,y,z)ωf3=fdxdydz.
We can use these correspondences to define various operations involving functions and vector fields. For example, suppose f : 3 f : 3 f:^(3)rarrf:{ }^{3} \rightarrowf:3 1 R 1 R 1R1 R1R is a 0 -form. Then d f d f dfd fdf is the 1 -form, f x d x + f y d y + f z d z f x d x + f y d y + f z d z (del f)/(del x)dx+(del f)/(del y)dy+(del f)/(del z)dz\frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y+\frac{\partial f}{\partial z} d zfxdx+fydy+fzdz. The vector field associated to this 1 -form is then f x , f y , f z . In calculus we call f x , f y , f z . In calculus we call  ^((:(del f)/(del x),(del f)/(del y),(del f)/(del z):)". In calculus we call "){ }^{\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle \text {. In calculus we call }}fx,fy,fz. In calculus we call  this vector field grad f f fff, or f f grad f\nabla ff. In other words, f f grad f\nabla ff is the vector field associated with the 1-form, d f d f dfd fdf. This can be summarized by the equation
d f = ω f f 1 d f = ω f f 1 df=omega_(grad ff)^(1)d f=\omega_{\nabla f f}^{1}df=ωff1
It will be useful to think of this as a diagram as well.
Example 38. Suppose f = x 2 y 3 z f = x 2 y 3 z f=x^(2)y^(3)zf=x^{2} y^{3} zf=x2y3z. Then d f = 2 x y 3 z d x + 3 x 2 y 2 z d y d f = 2 x y 3 z d x + 3 x 2 y 2 z d y df=2xy^(3)zdx+3x^(2)y^(2)zdyd f=2 x y^{3} z d x+3 x^{2} y^{2} z d ydf=2xy3zdx+3x2y2zdy + x 3 y 3 d z + x 3 y 3 d z +x^(3)y^(3)dz+x^{3} y^{3} d z+x3y3dz. The associated vector field, grad f grad f grad f\operatorname{grad} fgradf, is then f = 2 x y 3 z f = 2 x y 3 z grad f=(:2xy^(3)z:}\nabla f=\left\langle 2 x y^{3} z\right.f=2xy3z, 3 x 2 y 2 z , x 3 y 3 3 x 2 y 2 z , x 3 y 3 {:3x^(2)y^(2)z,x^(3)y^(3):)\left.3 x^{2} y^{2} z, x^{3} y^{3}\right\rangle3x2y2z,x3y3.
Similarly, if we start with a vector field, F F F\mathbf{F}F, form the associated 1form, ω F ω F omega_(F)\omega_{\mathbf{F}}ωF, differentiate it, and look at the corresponding vector field, then the result is called curl F F F\mathbf{F}F, or × F × F grad xxF\nabla \times \mathbf{F}×F. So, × F × F grad xxF\nabla \times \mathbf{F}×F is the vector field
associated with the 2 -form, d ω F 1 d ω F 1 domega_(F)^(1)d \omega_{\mathbf{F}}^{1}dωF1. This can be summarized by the equation
d ω F 1 = ω × F 2 . d ω F 1 = ω × F 2 . domega_(F)^(1)=omega_(grad xxF)^(2).d \omega_{\mathbf{F}}^{1}=\omega_{\nabla \times \mathbf{F}}^{2} .dωF1=ω×F2.
This can also be illustrated by the following diagram.
Example 39. Let F = x y , y z , x 2 F = x y , y z , x 2 F=(:xy,yz,x^(2):)\mathbf{F}=\left\langle x y, y z, x^{2}\right\rangleF=xy,yz,x2. The associated 1-form is then
ω F 1 = x y d x + y z d y + x 2 d z . ω F 1 = x y d x + y z d y + x 2 d z . omega_(F)^(1)=xydx+yzdy+x^(2)dz.\omega_{\mathbf{F}}^{1}=x y d x+y z d y+x^{2} d z .ωF1=xydx+yzdy+x2dz.
The derivative of this 1 -form is the 2 -form
d ω F 1 = y d y d z + 2 x d x d z x d x d y . d ω F 1 = y d y d z + 2 x d x d z x d x d y . domega_(F)^(1)=-ydy^^dz+2xdx^^dz-xdx^^dy.d \omega_{\mathrm{F}}^{1}=-y d y \wedge d z+2 x d x \wedge d z-x d x \wedge d y .dωF1=ydydz+2xdxdzxdxdy.
The vector field associated to this 2-form is curl F, which is
× F = y , 2 x , x . × F = y , 2 x , x . grad xxF=(:-y,-2x,-x:).\nabla \times \mathbf{F}=\langle-y,-2 x,-x\rangle .×F=y,2x,x.
Lastly, we can start with a vector field, F = F x , F y , F z F = F x , F y , F z F=(:F_(x,)F_(y,)F_(z):)\mathbf{F}=\left\langle F_{x,} F_{y,} F_{z}\right\rangleF=Fx,Fy,Fz, and then look at the 3-form, d ω F 2 = ( F x x + F y y + F z z ) d x d y d z (see d ω F 2 = F x x + F y y + F z z d x d y d z (see  domega_(F)^(2)=((delF_(x))/(del x)+(delF_(y))/(del y)+(delF_(z))/(del z))dx^^dy^^dz_((see )d \omega_{\mathbf{F}}^{2}=\left(\frac{\partial F_{x}}{\partial x}+\frac{\partial F_{y}}{\partial y}+\frac{\partial F_{z}}{\partial z}\right) d x \wedge d y \wedge d z_{\text {(see }}dωF2=(Fxx+Fyy+Fzz)dxdydz(see 
Problem 6.13). The function, F x x + F y y + F z z F x x + F y y + F z z (delF_(x))/(del x)+(delF_(y))/(del y)+(delF_(z))/(del z)\frac{\partial F_{x}}{\partial x}+\frac{\partial F_{y}}{\partial y}+\frac{\partial F_{z}}{\partial z}Fxx+Fyy+Fzz is called div F F F\mathbf{F}F, or grad\nabla. F F F\mathbf{F}F. This is summarized in the following equation and diagram.
d ω F 2 = ω V F 3 F div F ω F 2 d d ω F 2 d ω F 2 = ω V F 3 F  div  F ω F 2 d d ω F 2 {:[domega_(F)^(2)=omega_(V*F)^(3)],[Frarr"" div ""grad*F],[darr],[omega_(F)^(2)rarr_("d")^("longrightarrow")domega_(F)^(2)]:}\begin{gathered} d \omega_{\mathbf{F}}^{2}=\omega_{\mathrm{V} \cdot \mathbf{F}}^{3} \\ \mathbf{F} \xrightarrow{\text { div }} \nabla \cdot \mathbf{F} \\ \downarrow \\ \omega_{\mathbf{F}}^{2} \xrightarrow[d]{\longrightarrow} d \omega_{\mathbf{F}}^{2} \end{gathered}dωF2=ωVF3F div FωF2ddωF2
Example 40. Let F = x y , y z , x 2 F = x y , y z , x 2 F=(:xy,yz,x^(2):)\mathbf{F}=\left\langle x y, y z, x^{2}\right\rangleF=xy,yz,x2. The associated 2-form is then
ω F 2 = x y d y d z y z d x d z + x 2 d x d y . ω F 2 = x y d y d z y z d x d z + x 2 d x d y . omega_(F)^(2)=xydy^^dz-yzdx^^dz+x^(2)dx^^dy.\omega_{\mathrm{F}}^{2}=x y d y \wedge d z-y z d x \wedge d z+x^{2} d x \wedge d y .ωF2=xydydzyzdxdz+x2dxdy.
The derivative is the 3 -form
d ω F 2 = ( y + z ) d x d y d z . d ω F 2 = ( y + z ) d x d y d z . domega_(F)^(2)=(y+z)dx^^dy^^dz.d \omega_{\mathrm{F}}^{2}=(y+z) d x \wedge d y \wedge d z .dωF2=(y+z)dxdydz.
So div F div F divF\operatorname{div} \mathbf{F}divF is the function F = y + z F = y + z grad*F=y+z\nabla \cdot \mathbf{F}=y+zF=y+z.
Two important vector identities follow from the fact that for a differential form, ω ω omega\omegaω, calculating d ( a ω ) d ( a ω ) d(a omega)d(a \omega)d(aω) always yields zero (see Problem 6.9). For the first identity, consider the following diagram.
This shows that if f f fff is a 0 -form, then the vector field corresponding to d d f d d f ddfd d fddf is × ( f ) × ( f ) grad xx(grad f)\nabla \times(\nabla f)×(f). But d d f = 0 d d f = 0 ddf=0d d f=0ddf=0, so we conclude
× ( f ) = 0 × ( f ) = 0 grad xx(grad f)=0\nabla \times(\nabla f)=0×(f)=0
For the second identity, consider the diagram
This shows that if d d ω Fis 1 d d ω Fis  1 ddomega_("Fis ")^(1)d d \omega_{\text {Fis }}^{1}ddωFis 1 written as g d x d y d z g d x d y d z gdx^^dy^^dzg d x \wedge d y \wedge d zgdxdydz, then the function g g ggg is equal to ( × F ) ( × F ) grad*(grad xxF)\nabla \cdot(\nabla \times \mathbf{F})(×F). But d d ω 1 = 0 d d ω 1 = 0 ddomega^(1)=0d d \omega^{\mathbf{1}}=0ddω1=0, so we conclude
( × F ) = 0 . ( × F ) = 0 . grad*(grad xxF)=0.\nabla \cdot(\nabla \times \mathbf{F})=0 .(×F)=0.
In vector calculus we also learn how to integrate vector fields over parameterized curves (1-chains) and surfaces (2-chains). Suppose first that σ σ sigma\sigmaσ is some parameterized curve. Then we can integrate the component of F F F\mathbf{F}F which points in the direction of the tangent vectors to σ σ sigma\sigmaσ. This integral is usually denoted by σ F d s σ F d s int_(sigma)F*ds\int_{\sigma} \mathbf{F} \cdot d \mathbf{s}σFds, and its definition is precisely the same as the definition we learned here for ω F 1 ω F 1 intomega_(F)^(1)\int \omega_{\mathbf{F}}^{1}ωF1 precisely the same as the definition we learned here for σ σ sigma\sigmaσ. A special case of this integral arises when F = f F = f F=grad f\mathbf{F}=\nabla fF=f, for some function, f f fff . In this case, ω Fis just 1 d f ω Fis just  1 d f omega_("Fis just ")^(1)df\omega_{\text {Fis just }}^{1} d fωFis just 1df, so the definition of σ f . d s σ f . d s int_(sigma)grad f.ds\int_{\sigma} \nabla f . d \mathbf{s}σf.ds is the same as σ d f as σ d f int_(assigma)df\int_{\mathrm{as} \sigma} d fasσdf.
7.23. Let C C CCC be any curve in R 3 R 3 R^(3)R^{3}R3 from ( 0 , 0 , 0 ) ( 0 , 0 , 0 ) (0,0,0)(0,0,0)(0,0,0) to ( 1 , 1 , 1 ) ( 1 , 1 , 1 ) (1,1,1)(1,1,1)(1,1,1). Let F F FFF be the vector field y z , x z , x y y z , x z , x y (:yz,xz,xy:)\langle y z, x z, x y\rangleyz,xz,xy. Show *that C C CCC F d d *d\cdot dd s does not depend on C.
We also learn to integrate vector fields over parameterized surfaces. In this case, the quantity we integrate is the component of the vector field which is normal to the surface. This integral is often denoted bys F d S F d S F*dS\mathbf{F} \cdot d \mathbf{S}FdS. Its definition is precisely the same as that of s s int_(s)\int_{s}s (see Problems 4.23 and 4.24). A special case of this is when F = F = F=grad\mathbf{F}=\nablaF= × G × G xxG\times \mathbf{G}×G, for some vector field, G G G\mathbf{G}G. Then ω Gis 2 ω Gis  2 omega_("Gis ")^(2)\omega_{\text {Gis }}^{2}ωGis 2 just d ω G d ω G ^(domega_(G)){ }^{d \omega_{\mathbf{G}}}dωG, so we see that S ( × G ) d S S ( × G ) d S int_(S)(grad xxG)*dS\int_{S}(\nabla \times \mathbf{G}) \cdot d \mathbf{S}S(×G)dS must be the same ass G d ω G 1 G d ω G 1 int_(G)domega_(G)^(1)\int_{\mathbf{G}} d \omega_{\mathbf{G}}^{1}GdωG1.
The most basic thing to integrate ver a 3 -dimensional region (i.e., a 3-chain), Ω Ω Omega\OmegaΩ, in R 3 R 3 R^(3)\mathrm{R}^{3}R3 is a function f ( x , y , x ) f ( x , y , x ) f(x,y,x)f(x, y, x)f(x,y,x). In calculus we denote this integral as f d V f d V int fdV\int f d VfdV. Note that this is precisely the same ass f f ^(f){ }^{f}f. A special case is when f = F f = F f=grad*Ff=\nabla \cdot \mathbf{F}f=F, for some vector field F F F\mathbf{F}F. In this case ω F 2 ˙ ω F 2 ˙ intomega_(F)^(2)^(˙)\int \dot{\omega_{\mathbf{F}}^{2}}ωF2˙
differential forms as Ω Ω Omega\OmegaΩ
We summarize the equivalence between the integrals developed in vector calculus and various integrals of differential forms in Table 7.1.
Table 7.1. The equivalence between the integrals of vector calculus and differential forms.
Let us now apply the generalized Stokes' Theorem to various situations. First, we start with a parameterization, φ : [ a , b ] σ φ : [ a , b ] σ varphi:[a,b]rarr sigma sub\varphi:[a, b] \rightarrow \sigma \subsetφ:[a,b]σ R 3 R 3 R^(3)\mathrm{R}^{3}R3, of a curve in R 3 R 3 R^(3)\mathrm{R}^{3}R3, and a function, f : R 3 R f : R 3 R f:R^(3)rarrRf: \mathrm{R}^{3} \rightarrow \mathrm{R}f:R3R. Then we have
σ f d s σ d f = σ f = f ( ϕ ( b ) ) f ( ϕ ( a ) ) . σ f d s σ d f = σ f = f ( ϕ ( b ) ) f ( ϕ ( a ) ) . int_(sigma)grad f*ds-=int_(sigma)df=int_(del sigma)f=f(phi(b))-f(phi(a)).\int_{\sigma} \nabla f \cdot d \mathbf{s} \equiv \int_{\sigma} d f=\int_{\partial \sigma} f=f(\phi(b))-f(\phi(a)) .σfdsσdf=σf=f(ϕ(b))f(ϕ(a)).
This shows the independence of path of line integrals of gradient fields. We can use this to prove that a line integral of a gradient field over any simple closed curve is 0 , but for us there is an easier, direct proof, which again uses the generalized Stokes' Theorem. Suppose σ σ sigma\sigmaσ is a simple closed loop in R 3 R 3 R^(3)R^{3}R3 (i.e., σ = φ σ = φ del sigma=varphi\partial \sigma=\varphiσ=φ ). Then σ = D σ = D sigma=del D\sigma=\partial Dσ=D, for some 2-chain, D D DDD. We now have
σ f d s σ d f = D d d f = 0 . σ f d s σ d f = D d d f = 0 . int_(sigma)grad f*ds-=int_(sigma)df=int_(D)ddf=0.\int_{\sigma} \nabla f \cdot d \mathbf{s} \equiv \int_{\sigma} d f=\int_{D} d d f=0 .σfdsσdf=Dddf=0.
Now, suppose we have a vector field, F, and a parameterized surface, S S SSS. Yet another application of the generalized Stokes' Theorem yields
S F d s S ω F 1 = S d ω F 1 S ( × F ) d S . S F d s S ω F 1 = S d ω F 1 S ( × F ) d S . int_(del S)F*ds-=int_(del S)omega_(F)^(1)=int_(S)domega_(F)^(1)-=int_(S)(grad xxF)*dS.\int_{\partial S} \mathbf{F} \cdot d \mathbf{s} \equiv \int_{\partial S} \omega_{\mathbf{F}}^{1}=\int_{S} d \omega_{\mathbf{F}}^{1} \equiv \int_{S}(\nabla \times \mathbf{F}) \cdot d \mathbf{S} .SFdsSωF1=SdωF1S(×F)dS.
In vector calculus we call this equality "Stokes' theorem." In some sense, × F × F grad xxF\nabla \times \mathbf{F}×F measures the "twisting" of F F F\mathbf{F}F at points of S S SSS. So Stokes' theorem says that the net twisting of F F F\mathbf{F}F over all of S S SSS is the same as the amount F F F\mathbf{F}F circulates around S S del S\partial SS.
Example 41. Suppose we are faced with a problem phrased as: "Use Stokes' theorem to calculate C F d C F d CF*dC \mathbf{F} \cdot \mathbf{d}CFd, where C C CCC is the curve of intersection of the cylinder x 2 + y 2 = 1 x 2 + y 2 = 1 x^(2)+y^(2)=1x^{2}+y^{2}=1x2+y2=1 and the plane z = x + 1 z = x + 1 z=x+1z=x+1z=x+1, and F F F\mathbf{F}F is the vector field x 2 y , x y 2 , z 3 . " x 2 y , x y 2 , z 3 . " (:-x^(2)y,xy^(2),z^(3):)."\left\langle-x^{2} y, x y^{2}, z^{3}\right\rangle . "x2y,xy2,z3."
We will solve this problem by translating to the language of differential forms, and using the generalized Stokes' Theorem, instead. To begin, note that F d s = F d s = F*ds=\mathbf{F} \cdot d \mathbf{s}=Fds=
c ω F 1 , and ω F 1 = x 2 y d x + x y 2 d y + z 3 d z . c ω F 1 ,  and  ω F 1 = x 2 y d x + x y 2 d y + z 3 d z . int_(c)omega_(F)^(1)," and "omega_(F)^(1)=-x^(2)ydx+xy^(2)dy+z^(3)dz.\int_{c} \omega_{\mathbf{F}}^{1}, \text { and } \omega_{\mathbf{F}}^{1}=-x^{2} y d x+x y^{2} d y+z^{3} d z .cωF1, and ωF1=x2ydx+xy2dy+z3dz.
Now, to use the generalized Stokes' Theorem we will need to calculate
d ω F 1 = ( x 2 + y 2 ) d x d y d ω F 1 = x 2 + y 2 d x d y domega_(F)^(1)=(x^(2)+y^(2))dx^^dyd \omega_{\mathbf{F}}^{1}=\left(x^{2}+y^{2}\right) d x \wedge d ydωF1=(x2+y2)dxdy
Let D D DDD denote the subset of the plane z = x + 1 z = x + 1 z=x+1z=x+1z=x+1 bounded by C C CCC. Then D = C D = C del D=C\partial D=CD=C. Hence, by the generalized Stokes' Theorem we have
C ω F 1 = D d ω F 1 = D ( x 2 + y 2 ) d x d y C ω F 1 = D d ω F 1 = D x 2 + y 2 d x d y int_(C)omega_(F)^(1)=int_(D)domega_(F)^(1)=int_(D)(x^(2)+y^(2))dx^^dy\int_{C} \omega_{\mathbf{F}}^{1}=\int_{D} d \omega_{\mathbf{F}}^{1}=\int_{D}\left(x^{2}+y^{2}\right) d x \wedge d yCωF1=DdωF1=D(x2+y2)dxdy
The region D D DDD is parameterized by Ψ ( r , θ ) = ( r cos θ , r sin θ , r Ψ ( r , θ ) = ( r cos θ , r sin θ , r Psi(r,theta)=(r cos theta,r sin theta,r\Psi(r, \theta)=(r \cos \theta, r \sin \theta, rΨ(r,θ)=(rcosθ,rsinθ,r cos θ + 1 cos θ + 1 cos theta+1\cos \theta+1cosθ+1 ), where 0 r 1 0 r 1 0 <= r <= 10 \leq r \leq 10r1 and 0 θ 2 π 0 θ 2 π 0 <= theta <= 2pi0 \leq \theta \leq 2 \pi0θ2π. Using this one can ( x 2 + y 2 ) d x d y = π 2 x 2 + y 2 d x d y = π 2 int(x^(2)+y^(2))dx^^dy=(pi)/(2)\int\left(x^{2}+y^{2}\right) d x \wedge d y=\frac{\pi}{2}(x2+y2)dxdy=π2. (and should!) show that D D DDD
7.24. Let C C CCC be the square with sides ( x , ± 1 , 1 x , ± 1 , 1 x,+-1,1x, \pm 1,1x,±1,1 ), where 1 x 1 1 x 1 -1 <= x <= 1-1 \leq x \leq 11x1 and ( ± 1 , y , 1 ± 1 , y , 1 +-1,y,1\pm 1, y, 1±1,y,1 ), where 1 y 1 1 y 1 -1 <= y <= 1-1 \leq y \leq 11y1, with the indicated orientation (see Figure 7.5). Let F F F\mathbf{F}F be the vector field x y , x 2 , y 2 z x y , x 2 , y 2 z (:xy,x^(2),y^(2)z:)\left\langle x y, x^{2}, y^{2} z\right\ranglexy,x2,y2z. Computec F d s F d s F*ds\mathbf{F} \cdot d \mathbf{s}Fds.
Suppose now that Ω Ω Omega\OmegaΩ is some volume in R 3 R 3 R^(3)\mathrm{R}^{3}R3. Then we have
Fig. 7.5.
This last equality is called "Gauss' Divergence Theorem." F F grad*F\nabla \cdot \mathbf{F}F is a measure of how much F F F\mathbf{F}F "spreads out" at a point. So Gauss' theorem says that the total spreading out of F F F\mathbf{F}F inside Ω Ω Omega\OmegaΩ is the same as the net amount of F F F\mathbf{F}F "escaping" through Ω Ω del Omega\partial \OmegaΩ.
7.25. Let Ω Ω Omega\OmegaΩ be the cube { ( x , y , z ) 0 x , y , z 1 } { ( x , y , z ) 0 x , y , z 1 } {(x,y,z)∣0 <= x,y,z <= 1}\{(x, y, z) \mid 0 \leq x, y, z \leq 1\}{(x,y,z)0x,y,z1}. Let F F F\mathbf{F}F be the vector field x y 2 , y 3 , x 2 y 2 x y 2 , y 3 , x 2 y 2 (:xy^(2),y^(3),x^(2)y^(2):)\left\langle x y^{2}, y^{3}, x^{2} y^{2}\right\ranglexy2,y3,x2y2. Compute Ω F d S Ω F d S del OmegaF*dS\partial \Omega \mathbf{F} \cdot d \mathbf{S}ΩFdS. Ω Ω del Omega\partial \OmegaΩ
8

Applications

8.1 Maxwell's equations

As a brief application, we show how the language of differential forms can greatly simplify the classical vector equations of Maxwell. Much of this material is taken from [MTW73], where the interested student can find many more applications of differential forms to physics.
Maxwell's equations describe the relationship between electric and magnetic fields. Classically, both electricity and magnetism are described as a 3-dimensional vector field which varies with time:
E = E x , E y , E z B = B x , B y , B z , E = E x , E y , E z B = B x , B y , B z , {:[E=(:E_(x),E_(y),E_(z):)],[B=(:B_(x),B_(y),B_(z):)","]:}\begin{aligned} \mathbf{E} & =\left\langle E_{x}, E_{y}, E_{z}\right\rangle \\ \mathbf{B} & =\left\langle B_{x}, B_{y}, B_{z}\right\rangle, \end{aligned}E=Ex,Ey,EzB=Bx,By,Bz,
where E x , E z E z B x , B y E x , E z E z B x , B y E_(x),E_(z)E_(z)B_(x),B_(y)E_{x}, E_{z} E_{z} B_{x}, B_{y}Ex,EzEzBx,By, and B z B z B_(z)B_{z}Bz are all functions of x , y , z x , y , z x,y,zx, y, zx,y,z and t t ttt. Maxwell's equations are then:
B = 0 B t + × E = 0 E = 4 π ρ E t × B = 4 π J . B = 0 B t + × E = 0 E = 4 π ρ E t × B = 4 π J . {:[grad*B=0],[(delB)/(del t)+grad xxE=0],[grad*E=4pi rho],[(delE)/(del t)-grad xxB=-4piJ.]:}\begin{aligned} \nabla \cdot \mathbf{B} & =0 \\ \frac{\partial \mathbf{B}}{\partial t}+\nabla \times \mathbf{E} & =0 \\ \nabla \cdot \mathbf{E} & =4 \pi \rho \\ \frac{\partial \mathbf{E}}{\partial t}-\nabla \times \mathbf{B} & =-4 \pi \mathbf{J} . \end{aligned}B=0Bt+×E=0E=4πρEt×B=4πJ.
The quantity ρ ρ rho\rhoρ is called the charge density and the vector J = J x J = J x J=(:J_(x):}\mathbf{J}=\left\langle J_{x}\right.J=Jx J y J z J y J z {:J_(y^('))J_(z):)\left.J_{y^{\prime}} J_{z}\right\rangleJyJz is called the current density.
We can make all of this look much simpler by making the following definitions. First, we define a 2 -form called the Faraday, which
simultaneously describes both the electric and magnetic fields:
F = E x d x d t + E y d y d t + E z d z d t + B x d y d z + B y d z d x + B z d x d y F = E x d x d t + E y d y d t + E z d z d t + B x d y d z + B y d z d x + B z d x d y {:[F=E_(x)dx^^dt+E_(y)dy^^dt+E_(z)dz^^dt],[+B_(x)dy^^dz+B_(y)dz^^dx+B_(z)dx^^dy]:}\begin{aligned} \mathbf{F}= & E_{x} d x \wedge d t+E_{y} d y \wedge d t+E_{z} d z \wedge d t \\ & +B_{x} d y \wedge d z+B_{y} d z \wedge d x+B_{z} d x \wedge d y \end{aligned}F=Exdxdt+Eydydt+Ezdzdt+Bxdydz+Bydzdx+Bzdxdy
Next we define the "dual" 2-form, called the Maxwell:
F = E x d y d z + E y d z d x + E z d x d y + B x d t d x + B y d t d y + B z d t d z . F = E x d y d z + E y d z d x + E z d x d y + B x d t d x + B y d t d y + B z d t d z . {:[^(**)F=E_(x)dy^^dz+E_(y)dz^^dx+E_(z)dx^^dy],[+B_(x)dt^^dx+B_(y)dt^^dy+B_(z)dt^^dz.]:}\begin{aligned} { }^{*} \mathbf{F}= & E_{x} d y \wedge d z+E_{y} d z \wedge d x+E_{z} d x \wedge d y \\ & +B_{x} d t \wedge d x+B_{y} d t \wedge d y+B_{z} d t \wedge d z . \end{aligned}F=Exdydz+Eydzdx+Ezdxdy+Bxdtdx+Bydtdy+Bzdtdz.
We also define the 4-current, J J J\mathbf{J}J, and its "dual," J J ^(J){ }^{\mathbf{J}}J :
J = ρ , J x , J y , J z J = = ρ d x d y d z J x d t d y d z J y d t d z d x J z d t d x d y J = ρ , J x , J y , J z J = = ρ d x d y d z J x d t d y d z J y d t d z d x J z d t d x d y {:[J=(:rho,J_(x),J_(y),J_(z):)],[^(J=)=rho dx^^dy^^dz],[-J_(x)dt^^dy^^dz],[-J_(y)dt^^dz^^dx],[-J_(z)dt^^dx^^dy]:}\begin{aligned} \mathbf{J}= & \left\langle\rho, J_{x}, J_{y}, J_{z}\right\rangle \\ { }^{\mathbf{J}=}= & \rho d x \wedge d y \wedge d z \\ & -J_{x} d t \wedge d y \wedge d z \\ & -J_{y} d t \wedge d z \wedge d x \\ & -J_{z} d t \wedge d x \wedge d y \end{aligned}J=ρ,Jx,Jy,JzJ==ρdxdydzJxdtdydzJydtdzdxJzdtdxdy
Maxwell's four vector equations now reduce to:
d F = 0 d F = 4 π J . d F = 0 d F = 4 π J . {:[dF=0],[d^(**)F=4pi^(**)J.]:}\begin{aligned} d \mathbf{F} & =0 \\ d^{*} \mathbf{F} & =4 \pi^{*} \mathbf{J} . \end{aligned}dF=0dF=4πJ.
8.1. Show that the equation d F = 0 d F = 0 dF=0d \mathbf{F}=0dF=0 implies the first two of Maxwell's equations.
8.2. Show that the equation d F = 4 π d F = 4 π d_(**)F=4pi_(***)d_{*} \mathbf{F}=4 \pi_{\star}dF=4π J implies the second two of Maxwell's equations.
The differential form version of Maxwell's equation has a huge advantage over the vector formulation: it is coordinate free! A 2form such as F is an operator that "eats" pairs of vectors and "spits out" numbers. The way it acts is completely geometric ... that is, it
can be defined without any reference to the coordinate system ( t , x t , x t,xt, xt,x, y , z y , z y,zy, zy,z ). This is especially poignant when one realizes that Maxwell's equations are laws of nature that should not depend on a man-made construction such as coordinates.

8.2 Foliations and contact structures

Everyone has seen tree rings and layers in sedimentary rock. These are examples of foliations. Intuitively, a foliation is when some region of space has been "filled up" with lower-dimensional surfaces. A full treatment of foliations is a topic for a much larger textbook than this one. Here we will only be discussing foliations of R 3 R 3 R^(3)\mathrm{R}^{3}R3.
Let U U UUU be an open subset of R 3 R 3 R^(3)R^{3}R3. We say U U UUU has been foliated if there is a family φ t : R t U φ t : R t U varphi^(t):R_(t)rarr U\varphi^{t}: R_{t} \rightarrow Uφt:RtU of parameterizations (where for each t t ttt the domain R t R 2 R t R 2 R_(t)subR^(2)R_{t} \subset R^{2}RtR2 ) such that every point of U U UUU is in the image of exactly one such parameterization. In other words, the images of the parameterizations φ t φ t varphi^(t)\varphi^{t}φt are surfaces that fill up U U UUU, and no two overlap.
Suppose p p ppp is a point of U U UUU and U U UUU has been foliated as above. Then there is a unique value of t t ttt such that p p ppp is a point in φ t ( R t ) φ t R t varphi^(t)(R_(t))\varphi^{t}\left(R_{t}\right)φt(Rt). The partial derivatives, ϕ t x ( p ) ϕ t x ( p ) (delphi^(t))/(del x)(p)\frac{\partial \phi^{t}}{\partial x}(p)ϕtx(p) and ϕ t y ( p ) ϕ t y ( p ) (delphi^(t))/(del y)(p)\frac{\partial \phi^{t}}{\partial y}(p)ϕty(p) are then two vectors that span a plane in T p R 3 T p R 3 T_(p)R^(3)T_{p} \mathrm{R}^{3}TpR3. Let's call this plane Π p Π p Pi_(p)\Pi_{p}Πp. In other words, if U U UUU is foliated, then at every point p p ppp of U U UUU we get a plane Π p Π p Pi_(p)\Pi_{p}Πp in T p R 3 T p R 3 T_(p)R^(3)T_{p} \mathrm{R}^{3}TpR3.
The family { Π p } Π p {Pi_(p)}\left\{\Pi_{p}\right\}{Πp} is an example of a plane field. In general, a plane field is just a choice of a plane in each tangent space which varies smoothly from point to point in R 3 R 3 R^(3)\mathrm{R}^{3}R3. We say a plane field is integrable if it consists of the tangent planes to a foliation.
This should remind you a little of first-term calculus. If f : R 1 R 1 f : R 1 R 1 f:R^(1)rarrR^(1)f: \mathrm{R}^{1} \rightarrow \mathrm{R}^{1}f:R1R1 is a differentiable function, then at every point p p ppp on its graph we get a line in T p R 2 T p R 2 T_(p)R^(2)T_{p} R^{2}TpR2 (see Figure 4.2). If we only know the lines and want the original function, then we integrate.
There is a theorem that says that every line field on R 2 R 2 R^(2)\mathrm{R}^{2}R2 is integrable. The question we would like to answer in this section is whether or not this is true of plane fields on R 3 R 3 R^(3)\mathrm{R}^{3}R3. The first step is to figure out how to specify a plane field in some reasonably nice way. This is where differential forms come in. Suppose { Π p } Π p {Pi_(p)}\left\{\Pi_{p}\right\}{Πp} is a plane field. At each point p p ppp, we can define a line in T p R 3 T p R 3 T_(p)R^(3)T_{p} \mathrm{R}^{3}TpR3 (i.e., a line field) by looking at the set of all vectors that are perpendicular to Π p Π p Pi_(p)\Pi_{p}Πp. We can then define a 1-form ω ω omega\omegaω by projecting vectors onto these lines. So, if V p V p V_(p)V_{p}Vp is a vector in Π p Π p Pi_(p)\Pi_{p}Πp then ω ( V p ) = 0 ω V p = 0 omega(V_(p))=0\omega\left(V_{p}\right)=0ω(Vp)=0. Another way to say this is that the plane Π p Π p Pi_(p)\Pi_{p}Πp is the set of all vectors which yield zero when plugged into ω ω omega\omegaω. In shorthand, we write this set as Ker ω ω omega\omegaω ("Ker" comes from the word "Kernel," a term from linear algebra). So all we are saying is that ω ω omega\omegaω is a 1 -form such that Π p = Ker ω Π p = Ker ω Pi_(p)=Ker omega\Pi_{p}=\operatorname{Ker} \omegaΠp=Kerω. This is very convenient. To specify a plane field, all we have to do now is write down a 1-form!
Example 42. Suppose ω = d x ω = d x omega=dx\omega=d xω=dx. Then, at each point p p ppp of R 3 R 3 R^(3)\mathrm{R}^{3}R3, the vectors of T p R 3 T p R 3 T_(p)R^(3)T_{p} \mathrm{R}^{3}TpR3 that yield zero when plugged into ω ω omega\omegaω are all those in the d y d z d y d z dydzd y d zdydz-plane. Hence, Ker ω ω omega\omegaω is the plane field consisting of all of the d y d z d y d z dydzd y d zdydz-planes (one for every point of 3 3 ^(3){ }^{3}3 ). It is obvious that this plane field is integrable; at each point p p ppp we just have the tangent plane to the plane parallel to the y z y z yzy zyz-plane through p p ppp.
In the above example, note that any 1-form that looks like f ( x , y f ( x , y f(x,yf(x, yf(x,y, z ) d x z ) d x z)dxz) d xz)dx defines the same plane field, as long as f f fff is non-zero everywhere. So, knowing something about a plane field (like the assumption that it is integrable) seems like it might not say much about the 1 -form ω ω omega\omegaω, since so many different 1 -forms give the same plane field. Let's investigate this further.
First, let's see if there is anything special about the derivative of a 1-form that looks like ω = f ( x , y , z ) d x ω = f ( x , y , z ) d x omega=f(x,y,z)dx\omega=f(x, y, z) d xω=f(x,y,z)dx. This is easy: d ω = f y d y d x + f z d z d x d ω = f y d y d x + f z d z d x d omega=(del f)/(del y)dy^^dx+(del f)/(del z)dz^^dxd \omega=\frac{\partial f}{\partial y} d y \wedge d x+\frac{\partial f}{\partial z} d z \wedge d xdω=fydydx+fzdzdx. This is nothing special so far. What about combining this with ω ω omega\omegaω ? Let's compute:
ω d ω = f ( x , y , z ) d x ( f y d y d x + f z d z d x ) = 0 ω d ω = f ( x , y , z ) d x f y d y d x + f z d z d x = 0 omega^^d omega=f(x,y,z)dx^^((del f)/(del y)dy^^dx+(del f)/(del z)dz^^dx)=0\omega \wedge d \omega=f(x, y, z) d x \wedge\left(\frac{\partial f}{\partial y} d y \wedge d x+\frac{\partial f}{\partial z} d z \wedge d x\right)=0ωdω=f(x,y,z)dx(fydydx+fzdzdx)=0
Now that is special! In fact, recall our earlier emphasis on the fact that forms are coordinate free. In other words, any computation one can perform with forms will give the same answer regardless of what coordinates are chosen. The wonderful thing about foliations is that near every point you can always choose coordinates so that your foliation looks like planes parallel to the y z y z yzy zyz-plane. In other words, the above computation is not as special as you might think:
Theorem 2. If Ker ω ω omega\omegaω is an integrable plane field, then ω a ω = 0 ω a ω = 0 omega^^a omega=0\omega \wedge a \omega=0ωaω=0 at every point of R 3 R 3 R^(3)\mathrm{R}^{3}R3.
It should be noted that we have only chosen to work in R 3 R 3 R^(3)\mathrm{R}^{3}R3 for ease of visualization. There are higher-dimensional definitions of foliations and plane fields. In general, if the kernel of a 1-form ω ω omega\omegaω defines an integrable plane field then ω a ω n = 0 ω a ω n = 0 omega^^aomega^(n)=0\omega \wedge a \omega^{n}=0ωaωn=0.
Our search for a plane field that is not integrable (i.e., not the tangent planes to a foliation) has now been reduced to the search for a 1-form ω ω omega\omegaω for which ω a ω 0 ω a ω 0 omega^^a omega!=0\omega \wedge a \omega \neq 0ωaω0 somewhere. There are many such forms. An easy one is x d y + d z x d y + d z xdy+dzx d y+d zxdy+dz. We compute:
( x d y + d z ) d ( x d y + d z ) = ( x d y + d z ) ( d x d y ) = d z d x d y ( x d y + d z ) d ( x d y + d z ) = ( x d y + d z ) ( d x d y ) = d z d x d y (xdy+dz)^^d(xdy+dz)=(xdy+dz)^^(dx^^dy)=dz^^dx^^dy(x d y+d z) \wedge d(x d y+d z)=(x d y+d z) \wedge(d x \wedge d y)=d z \wedge d x \wedge d y(xdy+dz)d(xdy+dz)=(xdy+dz)(dxdy)=dzdxdy
Our answer is quite special. All we needed was a 1-form such that
ω d ω 0 ω d ω 0 omega^^d omega!=0\omega \wedge d \omega \neq 0ωdω0
somewhere. What we found was a 1-form for which ω a ω 0 ω a ω 0 omega^^a omega!=0\omega \wedge a \omega \neq 0ωaω0 everywhere. This means that there is not a single point of R 3 R 3 R^(3)\mathrm{R}^{3}R3 which has a neighborhood in which the planes given by Ker x d y + d z x d y + d z xdy+dzx d y+d zxdy+dz are tangent to a foliation. Such a plane field is called a contact structure.
At this point you are probably wondering, "What could Ker x d y x d y xdyx d yxdy + d z + d z +dz+d z+dz possibly look like?!" It is not so easy to visualize this, but we have tried to give you some indication in Figure 8.1.1 A good exercise is to stare at this picture long enough to convince yourself that the planes pictured cannot be the tangent planes to a foliation.
We have just seen how we can use differential forms to tell if a plane field is integrable. But one may still wonder if there is more we can say about a 1 -form, assuming its kernel is integrable. Let's go back to the expression ω a ω ω a ω omega^^a omega\omega \wedge a \omegaωaω. Recall that ω ω omega\omegaω is a 1 -form, which makes a ω a ω a omegaa \omegaaω a 2 -form, and hence ω a ω ω a ω omega^^a omega\omega \wedge a \omegaωaω a 3 -form.
A 3-form on T p R 3 T p R 3 T_(p)R^(3)T_{p} \mathrm{R}^{3}TpR3 measures the volume of the parallelepiped spanned by three vectors, multiplied by a constant. For example, if Ψ = a β γ Ψ = a β γ Psi=a^^beta^^gamma\Psi=a \wedge \beta \wedge \gammaΨ=aβγ is a 3 -form, then the constant it scales volume by is given by the volume of the parallelepiped spanned by the vectors a a (:a:)\langle a\ranglea , β β (:beta:)\langle\beta\rangleβ and γ γ gamma:)\gamma\rangleγ (where " a a (:a:)\langle a\ranglea " refers to the vector dual to the 1-form a a aaa introduced in Section 4.3). If it turns out that Ψ Ψ Psi\PsiΨ is the zero 3 -form, then the vector a a (:a:)\langle a\ranglea must be in the plane spanned by the vectors β β (:beta:)\langle\beta\rangleβ and γ γ (:gamma:)\langle\gamma\rangleγ.
Fig. 8.1. The plane field Ker x d y + d z Ker x d y + d z Ker xdy+dz\operatorname{Ker} x d y+d zKerxdy+dz.
On R 3 R 3 R^(3)\mathrm{R}^{3}R3 the results of Section 4.3 tell us that a 2 -form such as α ω α ω alpha omega\alpha \omegaαω can always be written as a β a β a^^betaa \wedge \betaaβ, for some 1 -forms a a aaa and β β beta\betaβ. If ω ω omega\omegaω is a 1 -form with integrable kernel, then we have already seen that ω d ω d omega^^d\omega \wedge dωd ω = ω a β = 0 ω = ω a β = 0 omega=omega^^a^^beta=0\omega=\omega \wedge a \wedge \beta=0ω=ωaβ=0. But this tells us that ω ω (:omega:)\langle\omega\rangleω must be in the plane spanned by the vectors a a (:a:)\langle a\ranglea and β β (:beta:)\langle\beta\rangleβ. Now we can invoke Lemma 1 of Chapter 4, which says that we can rewrite α ω α ω alpha omega\alpha \omegaαω as ω v ω v omega^^v\omega \wedge vωv, for some 1form V. (See also Problem 4.27.)
If we start with a foliation and choose a 1 -form ω ω omega\omegaω whose kernel consists of planes tangent to the foliation, then the 1 -form v v vvv that we have just found is in no way canonical. We made a lot of choices to get to V , and different choices will end up with different 1 -forms. But here is the amazing fact: the integral of the 3-form α α vv^^alpha vv\vee \wedge \alpha \veeα does not depend on any of our choices! It is completely determined by the original foliation. Whenever a mathematician runs into a situation like this they usually throw up their hands and say, "Eureka! I've discovered an invariant." The quantity a a int vv^^a vv\int \vee \wedge a \veea is referred to as the Gobillion-Vey invariant of the foliation. It is a top c of current
research to identify exactly what information this number tells us about the foliation.
Two special cases are worth noting. First, it may turn out that vv^^\vee \wedge a V = 0 a V = 0 aV=0a \mathrm{~V}=0a V=0 everywhere. This tells us that the plane field given by Ker v is integrable, so we get another foliation. The other interesting case is when a a vv^^a vv\vee \wedge a \veea is nowhere zero. Then we get a contact structure.

8.3 How not to visualize a differential 1-form

There are several contemporary physics texts that attempt to give a visual interpretation of differential forms that seems quite different from the one presented here. As this alternate interpretation is much simpler than anything described in these notes, one may wonder why we have not taken this approach.
Let's look again at the 1 -form d x d x dxd xdx on R 3 R 3 R^(3)\mathrm{R}^{3}R3. Given a vector V p V p V_(p)V_{p}Vp at a point p p ppp, the value of d x ( V p ) d x V p dx(V_(p))d x\left(V_{p}\right)dx(Vp) is just the projection of V p V p V_(p)V_{p}Vp onto the d x d x dxd xdx axis in T p R 3 T p R 3 T_(p)R^(3)T_{p} \mathrm{R}^{3}TpR3. Now, let C C CCC be some parameterized curve in R 3 R 3 R^(3)\mathrm{R}^{3}R3 for which the x x xxx-coordinate is always increasing. Then c d x c d x cdxc d xcdx is just the length of the projection of C C CCC onto the x x xxx-axis. To the nearest intege, this is just the number of planes that C C CCC punctures of the form x = n x = n x=nx=nx=n, where n n nnn is an integer. So one way to visualize the form d x d x dxd xdx is to picture these planes.
Fig. 8.2. "Surfaces" of ω ω omega\omegaω ?
This view is very appealing. After all, every 1 -form ω ω omega\omegaω, at every point p p ppp, projects vectors onto some line I p I p I_(p)I_{p}Ip. So can we integrate ω ω omega\omegaω along a curve C C CCC (at least to the nearest integer) by counting the number of surfaces punctured by C C CCC whose tangent planes are perpendicular to the lines I p I p I_(p)I_{p}Ip (see Figure 8.2)? If you have read the previous section, you might guess that the answer is a categorical NO!
Recall that the planes perpendicular to the lines I p I p I_(p)I_{p}Ip are precisely Ker ω ω omega\omegaω. To say that there are surfaces whose tangent planes are perpendicular to the lines I p I p I_(p)I_{p}Ip is the same thing as saying that Ker ω Ker ω Ker omega\operatorname{Ker} \omegaKerω is an integrable plane field. But we have seen in the previous section that there are 1 -forms as simple as x d y + d z x d y + d z xdy+dzx d y+d zxdy+dz whose kernels are nowhere integrable.
Fig. 8.3. The Reeb foliation of the solid torus.
Can we at least use this interpretation for a 1-form whose kernel is integrable? Unfortunately, the answer is still no. Let ω ω omega\omegaω be the 1form on the solid torus whose kernel consists of the planes tangent to the foliation pictured in Figure 8.3. (This is called the Reeb foliation of the solid torus.) The surfaces of this foliation spiral continually outward. So if we try to pick some number of "sample" surfaces, then they will "bunch up" near the boundary torus. This seems to indicate that if we want to integrate ω ω omega\omegaω over any path that cuts through the solid torus, then we should get an infinite answer,
since such a path would intersect our "sample" surfaces an infinite number of times. However, we can certainly find a 1 -form ω ω omega\omegaω for which this is not the case.
We do not want to end this section on such a down note. Although it is not valid in general to visualize a 1 -form as a sample collection of surfaces from a foliation, we can visualize it as a plane field. For example, Figure 8.1 is a pretty good depiction of the 1form x d y + d z x d y + d z xdy+dzx d y+d zxdy+dz. In this picture there are a few evenly spaced elements of its kernel, but this is enough. To get a rough idea of the value of C x d y + d z C x d y + d z Cxdy+dzC x d y+d zCxdy+dz we can just count the number of (transverse) intersections of the planes pictured with C C CCC. So, for example, if C C CCC is a curve whose tangents are always contained in one of these planes (a so-called Legendrian curve), thencx d y + d z d y + d z dy+dzd y+d zdy+dz will be zero. Inspection of the picture reveals that examples of such curves are the lines parallel to the x x xxx-axis.
8.3. Show that if C C CCC is a line parallel to the x x xxx-axis, then C x d y + d z = C x d y + d z = Cxdy+dz=C x d y+d z=Cxdy+dz= 0.

9

Manifolds

9.1 Pull-backs

Before moving on to defining forms in more general contexts, we need to introduce one more concept. Let's re-examine Equation 5.3:
M ω = ± R ω ϕ ( x 1 , , x n ) ( ϕ x 1 ( x 1 , x n ) , , ϕ x n ( x 1 , x n ) ) d x 1 d x n . M ω = ± R ω ϕ x 1 , , x n ϕ x 1 x 1 , x n , , ϕ x n x 1 , x n d x 1 d x n . int_(M)omega=+-int_(R)omega_(phi(x_(1),dots,x_(n)))((del phi)/(delx_(1))(x_(1),dotsx_(n)),dots,(del phi)/(delx_(n))(x_(1),dotsx_(n)))dx_(1)^^dots^^dx_(n).\int_{M} \omega= \pm \int_{R} \omega_{\phi\left(x_{1}, \ldots, x_{n}\right)}\left(\frac{\partial \phi}{\partial x_{1}}\left(x_{1}, \ldots x_{n}\right), \ldots, \frac{\partial \phi}{\partial x_{n}}\left(x_{1}, \ldots x_{n}\right)\right) d x_{1} \wedge \ldots \wedge d x_{n} .Mω=±Rωϕ(x1,,xn)(ϕx1(x1,xn),,ϕxn(x1,xn))dx1dxn.
The form in the integrand on the right was defined so as to integrate to give the same answer as the form on the left. This is what we would like to generalize. Suppose φ : R n R m φ : R n R m varphi:R^(n)rarrR^(m)\varphi: R^{n} \rightarrow R^{m}φ:RnRm is a parameterization, and ω ω omega\omegaω is a k k kkk-form on R m R m R^(m)\mathrm{R}^{m}Rm. We define the pull-back of ω ω omega\omegaω under φ φ varphi\varphiφ to be the form on R n R n R^(n)\mathrm{R}^{n}Rn which gives the same integral over any k k kkk-cell, σ σ sigma\sigmaσ, as ω ω omega\omegaω does when integrated over φ ( σ ) φ ( σ ) varphi(sigma)\varphi(\sigma)φ(σ). Following convention, we denote the pull-back of ω ω omega\omegaω under φ φ varphi\varphiφ as " φ ω φ ω varphi_(***)omega\varphi_{\star} \omegaφω."
So how do we decide how φ ω φ ω varphi_(**)omega\varphi_{*} \omegaφω acts on a k k kkk-tuple of vectors in T p R n T p R n T_(p)R^(n)T_{p} \mathrm{R}^{n}TpRn ? The trick is to use φ φ varphi\varphiφ to translate the vectors to a k k kkk-tuple in T φ T φ T varphiT \varphiTφ ( p ) R m ( p ) R m _((p))R^(m){ }_{(p)} \mathrm{R}^{m}(p)Rm, and then plug them into ω ω omega\omegaω. The matrix D φ D φ D varphiD \varphiDφ, whose columns are the partial derivatives of φ φ varphi\varphiφ, is an n × m n × m n xx mn \times mn×m matrix. This matrix acts on vectors in T p R n T p R n T_(p)R^(n)T_{p} R^{n}TpRn, and returns vectors in T φ ( p ) R m T φ ( p ) R m T varphi(p)R^(m)T \varphi(p) R^{m}Tφ(p)Rm. So, we define (see Figure 9.1):
ϕ ω ( V p 1 , , V p k ) = ω ( D ϕ ( V p 1 ) , , D ϕ ( V p k ) ) . ϕ ω V p 1 , , V p k = ω D ϕ V p 1 , , D ϕ V p k . phi^(**)omega(V_(p)^(1),dots,V_(p)^(k))=omega(D phi(V_(p)^(1)),dots,D phi(V_(p)^(k))).\phi^{*} \omega\left(V_{p}^{1}, \ldots, V_{p}^{k}\right)=\omega\left(D \phi\left(V_{p}^{1}\right), \ldots, D \phi\left(V_{p}^{k}\right)\right) .ϕω(Vp1,,Vpk)=ω(Dϕ(Vp1),,Dϕ(Vpk)).
Example 43. Suppose ω = y d x + z d y + x d z ω = y d x + z d y + x d z omega=ydx+zdy+xdz\omega=y d x+z d y+x d zω=ydx+zdy+xdz is a 1-form on R 3 R 3 R^(3)\mathrm{R}^{3}R3, and φ ( a , b ) = ( a + b , a b , a b ) φ ( a , b ) = ( a + b , a b , a b ) varphi(a,b)=(a+b,a-b,ab)\varphi(a, b)=(a+b, a-b, a b)φ(a,b)=(a+b,ab,ab) is a map from R 2 R 2 R^(2)\mathrm{R}^{2}R2 to R 3 R 3 R^(3)\mathrm{R}^{3}R3. Then φ ω φ ω varphi_(***)omega\varphi_{\star} \omegaφω will be a 1-form on R 2 R 2 R^(2)\mathrm{R}^{2}R2. To determine which one, we can examine how it acts on the vectors 1 , 0 ( a , b ) 1 , 0 ( a , b ) (:1,0:)_((a,b))\langle 1,0\rangle_{(a, b)}1,0(a,b) and 0 , 1 ( a , b ) 0 , 1 ( a , b ) (:0,1:)_((a,b))\langle 0,1\rangle_{(a, b)}0,1(a,b).
Fig. 9.1. Defining φ ω φ ω varphi_(***)omega\varphi_{\star} \omegaφω.
ϕ ω ( 1 , 0 ( a , b ) ) = ω ( D ϕ ( 1 , 0 ( a , b ) ) ) = ω ( [ 1 1 1 1 b a ] [ 1 0 ] ( a , b ) ) = ω ( 1 , 1 , b ( a + b , a b , a b ) ) = ( a b ) + a b + ( a + b ) b = a b + 2 a b + b 2 . ϕ ω 1 , 0 ( a , b ) = ω D ϕ 1 , 0 ( a , b ) = ω 1 1 1 1 b a 1 0 ( a , b ) = ω 1 , 1 , b ( a + b , a b , a b ) = ( a b ) + a b + ( a + b ) b = a b + 2 a b + b 2 . {:[phi^(**)omega((:1,0:)_((a,b)))=omega(D phi((:1,0:)_((a,b))))],[=omega([[1,1],[1,-1],[b,a]][[1],[0]]_((a,b)))],[=omega((:1,1,b:)_((a+b,a-b,ab)))],[=(a-b)+ab+(a+b)b],[=a-b+2ab+b^(2).]:}\begin{aligned} \phi^{*} \omega\left(\langle 1,0\rangle_{(a, b)}\right) & =\omega\left(D \phi\left(\langle 1,0\rangle_{(a, b)}\right)\right) \\ & =\omega\left(\left[\begin{array}{rr} 1 & 1 \\ 1 & -1 \\ b & a \end{array}\right]\left[\begin{array}{l} 1 \\ 0 \end{array}\right]_{(a, b)}\right) \\ & =\omega\left(\langle 1,1, b\rangle_{(a+b, a-b, a b)}\right) \\ & =(a-b)+a b+(a+b) b \\ & =a-b+2 a b+b^{2} . \end{aligned}ϕω(1,0(a,b))=ω(Dϕ(1,0(a,b)))=ω([1111ba][10](a,b))=ω(1,1,b(a+b,ab,ab))=(ab)+ab+(a+b)b=ab+2ab+b2.
Similarly,
ϕ ω ( 0 , 1 ( a , b ) ) = ω ( 1 , 1 , a ( a + b , a b , a b ) ) = ( a b ) a b + ( a + b ) a = a b + a 2 . ϕ ω 0 , 1 ( a , b ) = ω 1 , 1 , a ( a + b , a b , a b ) = ( a b ) a b + ( a + b ) a = a b + a 2 . {:[phi^(**)omega((:0,1:)_((a,b)))=omega((:1,-1,a:)_((a+b,a-b,ab)))],[=(a-b)-ab+(a+b)a],[=a-b+a^(2).]:}\begin{aligned} \phi^{*} \omega\left(\langle 0,1\rangle_{(a, b)}\right) & =\omega\left(\langle 1,-1, a\rangle_{(a+b, a-b, a b)}\right) \\ & =(a-b)-a b+(a+b) a \\ & =a-b+a^{2} . \end{aligned}ϕω(0,1(a,b))=ω(1,1,a(a+b,ab,ab))=(ab)ab+(a+b)a=ab+a2.
Hence,
ϕ ω = ( a b + 2 a b + b 2 ) d a + ( a b + a 2 ) d b . ϕ ω = a b + 2 a b + b 2 d a + a b + a 2 d b . phi^(**)omega=(a-b+2ab+b^(2))da+(a-b+a^(2))db.\phi^{*} \omega=\left(a-b+2 a b+b^{2}\right) d a+\left(a-b+a^{2}\right) d b .ϕω=(ab+2ab+b2)da+(ab+a2)db.
9.1. If ω = x 2 d y d z + y 2 d z d w ω = x 2 d y d z + y 2 d z d w omega=x^(2)dy^^dz+y^(2)dz^^dw\omega=x^{2} d y \wedge d z+y^{2} d z \wedge d wω=x2dydz+y2dzdw is a 2-form on R 4 R 4 R^(4)\mathrm{R}^{4}R4, and φ ( a , b , c ) φ ( a , b , c ) varphi(a,b,c)\varphi(a, b, c)φ(a,b,c) = ( a , b , c , a b c ) = ( a , b , c , a b c ) =(a,b,c,abc)=(a, b, c, a b c)=(a,b,c,abc), then what is φ ω φ ω varphi_(***)omega\varphi_{\star} \omegaφω ?
9.2. If ω ω omega\omegaω is an n n nnn-form on R m R m R^(m)\mathrm{R}^{m}Rm and φ : R n R m φ : R n R m varphi:R^(n)rarrR^(m)\varphi: \mathrm{R}^{n} \rightarrow \mathrm{R}^{m}φ:RnRm, then
ϕ ω = ω ϕ ( x 1 , , x n ) ( ϕ x 1 ( x 1 , x n ) , , ϕ x n ( x 1 , x n ) ) d x 1 d x n ϕ ω = ω ϕ x 1 , , x n ϕ x 1 x 1 , x n , , ϕ x n x 1 , x n d x 1 d x n phi^(**)omega=omega_(phi(x_(1),dots,x_(n)))((del phi)/(delx_(1))(x_(1),dotsx_(n)),dots,(del phi)/(delx_(n))(x_(1),dotsx_(n)))dx_(1)^^dots^^dx_(n)\phi^{*} \omega=\omega_{\phi\left(x_{1}, \ldots, x_{n}\right)}\left(\frac{\partial \phi}{\partial x_{1}}\left(x_{1}, \ldots x_{n}\right), \ldots, \frac{\partial \phi}{\partial x_{n}}\left(x_{1}, \ldots x_{n}\right)\right) d x_{1} \wedge \ldots \wedge d x_{n}ϕω=ωϕ(x1,,xn)(ϕx1(x1,xn),,ϕxn(x1,xn))dx1dxn
In light of the preceding exercise, Equation 5.3 can be re-written as
M ω = R ϕ ω . M ω = R ϕ ω . int_(M)omega=int_(R)phi^(**)omega.\int_{M} \omega=\int_{R} \phi^{*} \omega .Mω=Rϕω.
9.3. If σ σ sigma\sigmaσ is a k k kkk-cell in R n , φ : R n R m R n , φ : R n R m R^(n),varphi:R^(n)rarrR^(m)\mathrm{R}^{n}, \varphi: \mathrm{R}^{n} \rightarrow \mathrm{R}^{m}Rn,φ:RnRm, and ω ω omega\omegaω is a k k kkk-form on R m R m R^(m)\mathrm{R}^{m}Rm then
σ ϕ ω = ϕ ( σ ) ω . σ ϕ ω = ϕ ( σ ) ω . int_(sigma)phi^(**)omega=int_(phi(sigma))omega.\int_{\sigma} \phi^{*} \omega=\int_{\phi(\sigma)} \omega .σϕω=ϕ(σ)ω.
9.4. If φ : R n R m φ : R n R m varphi:R^(n)rarrR^(m)\varphi: \mathrm{R}^{n} \rightarrow \mathrm{R}^{m}φ:RnRm and ω ω omega\omegaω is a k k kkk-form on R m R m R^(m)\mathrm{R}^{m}Rm, then d ( φ ω ) = φ ( a ω d φ ω = φ ( a ω d(varphi^(***)omega)=varphi^(**)(a omegad\left(\varphi^{\star} \omega\right)=\varphi^{*}(a \omegad(φω)=φ(aω ).
These exercises prepare us for the proof of the generalized Stokes' Theorem (recall that in Chapter 7 we only proved this theorem when
integrating over cubes and their boundaries). Suppose σ σ sigma\sigmaσ is an n n nnn-cell in R m , φ : I n R n R m R m , φ : I n R n R m R^(m),varphi:I^(n)subR^(n)rarrR^(m)\mathrm{R}^{m}, \varphi: I^{n} \subset \mathrm{R}^{n} \rightarrow \mathrm{R}^{m}Rm,φ:InRnRm is a parameterization of σ σ sigma\sigmaσ and ω ω omega\omegaω is an ( n n n-n-n 1 ) 1 ) 1)1)1)-form on R m R m R^(m)\mathrm{R}^{m}Rm. Then we can combine Problems 9.3, 9.4, and 7.7 to give us
σ ω = ϕ ( I n ) ω = I n ϕ ω = I n d ( ϕ ω ) = I n ϕ ( d ω ) = ϕ ( I n ) d ω = σ d ω σ ω = ϕ I n ω = I n ϕ ω = I n d ϕ ω = I n ϕ ( d ω ) = ϕ I n d ω = σ d ω int_(del sigma)omega=int_(phi(delI^(n)))omega=int_(delI^(n))phi^(**)omega=int_(I^(n))d(phi^(**)omega)=int_(I^(n))phi^(**)(d omega)=int_(phi(I^(n)))d omega=int_(sigma)d omega\int_{\partial \sigma} \omega=\int_{\phi\left(\partial I^{n}\right)} \omega=\int_{\partial I^{n}} \phi^{*} \omega=\int_{I^{n}} d\left(\phi^{*} \omega\right)=\int_{I^{n}} \phi^{*}(d \omega)=\int_{\phi\left(I^{n}\right)} d \omega=\int_{\sigma} d \omegaσω=ϕ(In)ω=Inϕω=Ind(ϕω)=Inϕ(dω)=ϕ(In)dω=σdω

9.2 Forms on subsets of R n R n R^(n)\mathrm{R}^{\boldsymbol{n}}Rn

The goal of this chapter is to slowly work up to defining forms in a more general setting than just on R n R n R^(n)\mathrm{R}^{n}Rn. One reason for this is because the generalized Stokes' Theorem actually tells us that forms on R n R n R^(n)\mathrm{R}^{n}Rn are not very interesting. For example, let's examine how a 1 -form, ω ω omega\omegaω , on R 2 R 2 R^(2)R^{2}R2, for which a ω = 0 a ω = 0 a omega=0a \omega=0aω=0 (i.e., ω ω omega\omegaω is closed), integrates over any 1chain, C C CCC, such that C = C = del C=O/\partial C=\varnothingC= (i.e., C C CCC is closed). It is a basic result of topology that any such 1 -chain bounds a 2-chain, D D DDD. Hence, c ω = c ω = c omega=c \omega=cω= D d ω = 0 D d ω = 0 int_(D)d omega=0\int_{D} d \omega=0Ddω=0 !
Fortunately, there is no reason to restrict ourselves to differential forms which are defined on all of R n R n R^(n)\mathrm{R}^{n}Rn. Instead, we can simply consider forms which are defined on subsets, U U UUU, of R n R n R^(n)R^{n}Rn. For technical reasons, we will always assume such subsets are open. This is a technical condition which means that for each p U p U p in Up \in UpU, there is an ϵ ϵ epsilon\epsilonϵ such that
{ q R n d ( p , q ) < ϵ } U . q R n d ( p , q ) < ϵ U . {q inR^(n)∣d(p,q) < epsilon}sub U.\left\{q \in \mathbb{R}^{n} \mid d(p, q)<\epsilon\right\} \subset U .{qRnd(p,q)<ϵ}U.
In this case, T U p = T R p n T U p = T R p n TU_(p)=TR_(p)^(n)T U_{p}=T R_{p}^{n}TUp=TRpn. Since a differential n n nnn-form is nothing more than a choice of an n n nnn-form on R n p R n p R^(n)_(p)\mathbb{R}^{n}{ }_{p}Rnp, for each p p ppp (with some condition about differentiability), it makes sense to talk about a differential form on U U UUU.
Example 44. is a differential 1 -form on R 2 ( 0 , 0 ) R 2 ( 0 , 0 ) R^(2)-(0,0)\mathrm{R}^{2}-(0,0)R2(0,0).
ω 0 = y x 2 + y 2 d x + x x 2 + y 2 d y ω 0 = y x 2 + y 2 d x + x x 2 + y 2 d y omega_(0)=-(y)/(x^(2)+y^(2))dx+(x)/(x^(2)+y^(2))dy\omega_{0}=-\frac{y}{x^{2}+y^{2}} d x+\frac{x}{x^{2}+y^{2}} d yω0=yx2+y2dx+xx2+y2dy
9.5. Show that a ω 0 = 0 a ω 0 = 0 aomega_(0)=0a \omega_{0}=0aω0=0.
9.6. Let C C CCC be the unit circle, oriented counter-clockwise. Show that C C CCC wo = 2 π = 2 π =2pi=2 \pi=2π. Hint: Let ω = y d x + x d y ω = y d x + x d y omega^(')=-ydx+xdy\omega^{\prime}=-y d x+x d yω=ydx+xdy. Note that on C , ω 0 = ω C , ω 0 = ω C,omega_(0)=omega^(')C, \omega_{0}=\omega^{\prime}C,ω0=ω.
If C C CCC is any closed 1 -chain in R 2 ( 0 , 0 ) R 2 ( 0 , 0 ) R^(2)-(0,0)R^{2}-(0,0)R2(0,0), then the quantity 1 2 π C is 1 2 π C  is  (1)/(2pi)int_(C" is ")\frac{1}{2 \pi} \int_{C \text { is }}12πC is  called the winding number of C C CCC, since it computes the number of times C C CCC winds around the origin.
9.7. Let x + x + x^(+)x^{+}x+denote the positive x x xxx-axis in R 2 ( 0 , 0 ) R 2 ( 0 , 0 ) R^(2)-(0,0)\mathrm{R}^{2}-(0,0)R2(0,0), and let C C CCC be any closed 1-chain. Suppose V p V p V_(p)V_{p}Vp is a basis vector of T C p T C p TC_(p)T C_{p}TCp which agrees with the orientation of C C CCC at p p ppp. A positive (respectively, negative) intersection of C C CCC with x + x + x^(+)x^{+}x+is one where V p V p V_(p)V_{p}Vp has a component which points "up" (respectively, "down"). Assume all intersections of C C CCC with x + x + x^(+)x^{+}x+are either positive or negative. Let P P PPP denote the number of positive ones and N N NNN the number of negative ones. Show that 1 2 π C ω 0 = P N 1 2 π C ω 0 = P N (1)/(2pi)int_(C)omega_(0)=P-N\frac{1}{2 \pi} \int_{C} \omega_{0}=P-N12πCω0=PN.
Hint: Use the generalized Stokes' Theorem.

9.3 Forms on parameterized subsets

Recall that at each point, a differential from is simply an alternating, multi-linear map on a tangent plane. So all we need to define a differential form on a more general space is a well-defined tangent space. One case in which this happens is when we have a parameterized subset of R m R m R^(m)\mathrm{R}^{m}Rm. Let φ : U R n M R m φ : U R n M R m varphi:U subR^(n)rarr M subR^(m)\varphi: U \subset \mathrm{R}^{n} \rightarrow M \subset \mathrm{R}^{m}φ:URnMRm be a (one-toone) parameterization of M M MMM. Then recall that T M p T M p TM_(p)T M_{p}TMp is defined to be the span of the partial derivatives of φ φ varphi\varphiφ at φ 1 ( p ) φ 1 ( p ) varphi^(-1)(p)\varphi^{-1}(p)φ1(p), and is a n n nnn-dimensional Euclidean space, regardless of the point, p p ppp. Hence, we say the dimension of M M MMM is n n nnn.
A differential k k kkk-form on M M MMM is simply an alternating, multilinear, real-valued function on T M p T M p TM_(p)T M_{p}TMp for each p M p M p in Mp \in MpM, which varies differentiably with p p ppp. In other words, a differential k k kkk-form on M M MMM is a whole family of k k kkk-forms, each one acting on T M p T M p TM_(p)T M_{p}TMp, for different points, p p ppp. It is not so easy to say precisely what we mean when we say the form varies in a differentiable way with p p ppp. Fortunately, we have already introduced the tools necessary to do this. Let's say that ω ω omega\omegaω is a family of k k kkk-forms, defined on T M p T M p TM_(p')T M_{p \prime}TMp for each p M p M p in Mp \in MpM. Then φ ω φ ω varphi_(**)omega\varphi_{*} \omegaφω is a family of k k kkk-forms, defined on TR n φ 1 ( p ) TR n φ 1 ( p ) TR^(n)varphi_(-1(p))\operatorname{TR}^{n} \varphi_{-1(p)}TRnφ1(p), for each p M p M p in Mp \in MpM. We say that ω ω omega\omegaω is a differentiable k k kkk-form on M M MMM, if φ ω φ ω varphi_(***)omega\varphi_{\star} \omegaφω is a differentiable family on U U UUU.
This definition illustrates an important technique which is often used when dealing with differential forms on manifolds. Rather than working in M M MMM directly, we use the map φ φ varphi_(**)\varphi_{*}φ to translate problems about forms on M M MMM into problems about forms on U U UUU. These are nice because we already know how to work with forms which are defined
on open subsets of R n R n R^(n)R^{n}Rn. We will have much more to say about this later.
Example 45. The infinitely long cylinder, L L LLL, of radius one, centered along the z z zzz - axis, is given by the parameterization, ϕ ( a , b ) = ( a a 2 + b 2 , b a 2 + b 2 , ln a 2 + b 2 ) ϕ ( a , b ) = a a 2 + b 2 , b a 2 + b 2 , ln a 2 + b 2 phi(a,b)=((a)/(sqrt(a^(2)+b^(2))),(b)/(sqrt(a^(2)+b^(2))),ln sqrt(a^(2)+b^(2)))\phi(a, b)=\left(\frac{a}{\sqrt{a^{2}+b^{2}}}, \frac{b}{\sqrt{a^{2}+b^{2}}}, \ln \sqrt{a^{2}+b^{2}}\right)ϕ(a,b)=(aa2+b2,ba2+b2,lna2+b2), whose domain is R 2 ( 0 R 2 ( 0 R^(2)-(0\mathrm{R}^{2}-(0R2(0, 0 ). We can use φ φ varphi_(***)\varphi_{\star}φ to solve any problem about forms on L L LLL, by translating it back to a problem about forms on U U UUU.
9.8. Consider the 1 -form, T = y d x + x d y T = y d x + x d y T^(')=-ydx+xdy\mathrm{T}^{\prime}=-y d x+x d yT=ydx+xdy, on R 3 R 3 R^(3)\mathrm{R}^{3}R3. In particular, this form acts on vectors in T L p T L p TL_(p)T L_{p}TLp, where L L LLL is the cylinder of the previous example, and p p ppp is any point in L L LLL. Let T T TTT be the restriction of T T T^(')T^{\prime}T to vectors in T L p T L p TL_(p)T L_{p}TLp. So, T is a 1 -form on L L LLL. Compute φ T φ T varphi_(***)T\varphi_{\star} \mathrm{T}φT. What does this tell you that T measures?
If ω ω omega\omegaω is a k k kkk-form on M M MMM, then what do we mean by d ω d ω d omegad \omegadω ? Whatever the definition, we clearly want a φ ω = φ a ω a φ ω = φ a ω avarphi_(***)omega=varphi_(***)a omegaa \varphi_{\star} \omega=\varphi_{\star} a \omegaaφω=φaω. So why do we not use this to define a ω a ω a omegaa \omegaaω ? After all, we know what a φ ω a φ ω avarphi_(***)omegaa \varphi_{\star} \omegaaφω is, since φ ω φ ω varphi_(***)omega\varphi_{\star} \omegaφω is a form on R n R n R^(n)\mathrm{R}^{n}Rn. Recall that D φ p D φ p Dvarphi_(p)D \varphi_{p}Dφp is a map from R n p R n p R^(n)_(p)\mathrm{R}^{n}{ }_{p}Rnp to R m p R m p R^(m)_(p)\mathrm{R}^{m}{ }_{p}Rmp. However, if we restrict the range to T M p T M p TM_(p)T M_{p}TMp, then D φ p D φ p Dvarphi_(p)D \varphi_{p}Dφp is one-to-one, so it makes sense to refer to D φ 1 D φ 1 Dvarphi^(-1)D \varphi^{-1}Dφ1. We now define
d ω ( V p 1 , , V p k + 1 ) = d ϕ ω ( D ϕ p 1 ( V p 1 ) , , D ϕ p 1 ( V p k + 1 ) ) . d ω V p 1 , , V p k + 1 = d ϕ ω D ϕ p 1 V p 1 , , D ϕ p 1 V p k + 1 . d omega(V_(p)^(1),dots,V_(p)^(k+1))=dphi^(**)omega(Dphi_(p)^(-1)(V_(p)^(1)),dots,Dphi_(p)^(-1)(V_(p)^(k+1))).d \omega\left(V_{p}^{1}, \ldots, V_{p}^{k+1}\right)=d \phi^{*} \omega\left(D \phi_{p}^{-1}\left(V_{p}^{1}\right), \ldots, D \phi_{p}^{-1}\left(V_{p}^{k+1}\right)\right) .dω(Vp1,,Vpk+1)=dϕω(Dϕp1(Vp1),,Dϕp1(Vpk+1)).
9.9. If T T T^(')T^{\prime}T and T T TTT are the 1-forms on R 3 R 3 R^(3)R^{3}R3 and L L LLL, respectively, defined in the previous section, compute d T d T dT^(')d T^{\prime}dT and a a aaa.

9.4 Forms on quotients of R 1 7 R 1 7 R^(17)\mathbf{R}^{\boldsymbol{1 7}}R17 (optional)

This section requires some knowledge of topology and algebra. It is not essential for the flow of the text.
While we are on the subject of differential forms on subsets of R n R n R^(n)\mathrm{R}^{n}Rn, there is a very common construction of a topological space for which it is very easy to define what we mean by a differential form. Let's look again at the cylinder, L L LLL, of the previous section. One way to construct L L LLL is to start with the plane, R 2 R 2 R^(2)\mathrm{R}^{2}R2, and "roll it up." More technically, we can consider the map, μ ( θ , z ) = ( cos θ , sin θ , z ) μ ( θ , z ) = ( cos θ , sin θ , z ) mu(theta,z)=(cos theta,sin theta,z)\mu(\theta, z)=(\cos \theta, \sin \theta, z)μ(θ,z)=(cosθ,sinθ,z). In general, this is a many-to-one map, so it is not a parameterization, in the strict sense. To remedy this, one might try and restrict the domain of μ μ mu\muμ to { ( θ , z ) R 2 0 θ < 2 π } ( θ , z ) R 2 0 θ < 2 π {(theta,z)inR^(2)∣0 <= theta < 2pi}\left\{(\theta, z) \in R^{2} \mid 0 \leq \theta<2 \pi\right\}{(θ,z)R20θ<2π}, however this set is not open.
Note that for each point, ( θ , z ) R 2 , D μ ( θ , z ) R 2 , D μ (theta,z)inR^(2),D mu(\theta, z) \in R^{2}, D \mu(θ,z)R2,Dμ is a one-to-one map from T R 2 θ , z T R 2 θ , z TR^(2)theta_(,z)T R^{2} \theta_{, z}TR2θ,z to T L μ ( θ z ) T L μ θ z TL_(mu)(theta_(z))T L_{\mu}\left(\theta_{z}\right)TLμ(θz). This is all we need in order for μ T μ T mu_(**)T\mu_{*} TμT to make sense, where T is the form on L L LLL defined in the previous section.
9.10. Show that μ T = θ μ T = θ mu_(**)T=O/ theta\mu_{*} \mathrm{~T}=\varnothing \thetaμ T=θ.
In this case, we say that μ μ mu\muμ is a covering map, R 2 R 2 R^(2)\mathrm{R}^{2}R2 is a cover of L L LLL, and d θ d θ d thetad \thetadθ is the lift of T to R 2 R 2 R^(2)\mathrm{R}^{2}R2.
9.11. Suppose ω 0 ω 0 omega_(0)\omega_{0}ω0 is the 1 -form on R 2 R 2 R^(2)\mathrm{R}^{2}R2 which we used to define the winding number. Let μ ( r , θ ) = ( r cos θ , r sin θ ) μ ( r , θ ) = ( r cos θ , r sin θ ) mu(r,theta)=(r cos theta,r sin theta)\mu(r, \theta)=(r \cos \theta, r \sin \theta)μ(r,θ)=(rcosθ,rsinθ). Let U = { ( r , θ ) r > U = { ( r , θ ) r > U={(r,theta)∣r >U=\{(r, \theta) \mid r>U={(r,θ)r> 0 } 0 } 0}0\}0}. Then μ : U { R 2 ( 0 , 0 ) } μ : U R 2 ( 0 , 0 ) mu:U rarr{R^(2)-(0,0)}\mu: U \rightarrow\left\{R^{2}-(0,0)\right\}μ:U{R2(0,0)} is a covering map. Hence, there is a one-to-one correspondence between a quotient of U U UUU and R 2 ( 0 , 0 ) R 2 ( 0 , 0 ) R^(2)-(0,0)R^{2}-(0,0)R2(0,0). Compute the lift of ω 0 ω 0 omega_(0)\omega_{0}ω0 to U U UUU.
Let's go back to the cylinder, L. Another way to look at things is to ask: How can we recover L L LLL from the θ z θ z theta z\theta zθz-plane? The answer is to
view L L LLL as a quotient space. Let's put an equivalence relation, R R RRR, on the points of R 2 R 2 R^(2)\mathrm{R}^{2}R2 : ( θ 1 , z 1 ) ( θ 2 , z 2 ) θ 1 , z 1 θ 2 , z 2 (theta_(1),z_(1))∼(theta_(2),z_(2))\left(\theta_{1}, z_{1}\right) \sim\left(\theta_{2}, z_{2}\right)(θ1,z1)(θ2,z2) if and only if z 1 = z 2 z 1 = z 2 z_(1)=z_(2)z_{1}=z_{2}z1=z2, and θ 1 θ 2 θ 1 θ 2 theta_(1)-theta_(2)\theta_{1}-\theta_{2}θ1θ2 = 2 n π = 2 n π =2n pi=2 n \pi=2nπ, for some n Z n Z n in Zn \in ZnZ. We will denote the quotient of R 2 R 2 R^(2)R^{2}R2 under this relation as R 2 / R R 2 / R R^(2)//R\mathrm{R}^{2} / RR2/R. μ μ mu\muμ now induces a one-to-one map, μ μ mu\muμ, from R 2 / R R 2 / R R^(2)//R\mathrm{R}^{2} / RR2/R onto L L LLL. Hence, these two spaces are homeomorphic.
Let's suppose now that we have a form on U U UUU, an open subset of R n R n R^(n)\mathrm{R}^{n}Rn, and we would like to know when it descends to a form on a quotient of U U UUU. Clearly, if we begin with the lift of a form, then it will descend. Let's try and see why. In general, if μ : U R n M R m μ : U R n M R m mu:U subR^(n)rarr M subR^(m)\mu: U \subset R^{n} \rightarrow M \subset R^{m}μ:URnMRm is a many-to-one map, differentiable at each point of U U UUU, then the sets, { μ 1 ( p ) } μ 1 ( p ) {mu^(-1)(p)}\left\{\mu^{-1}(p)\right\}{μ1(p)}, partition U U UUU. Hence, we can form the quotient space, U / μ 1 U / μ 1 U//mu^(-1)U / \mu^{-1}U/μ1, under this partition. For each x μ 1 ( p ) , D μ x x μ 1 ( p ) , D μ x x inmu^(-1)(p),Dmu_(x)x \in \mu^{-1}(p), D \mu_{x}xμ1(p),Dμx is a one-to-one map from T U x T U x TU_(x)T U_{x}TUx to T M p T M p TM_(p')T M_{p \prime}TMp and hence, D μ 1 D μ 1 Dmu^(-1)D \mu^{-1}Dμ1 is well-defined. If x x xxx and y y yyy are both in μ 1 ( p ) μ 1 ( p ) mu^(-1)(p)\mu^{-1}(p)μ1(p), then D μ 1 y = D μ x D μ 1 y = D μ x Dmu^(-1)y^(=)Dmu_(x)D \mu^{-1} y^{=} D \mu_{x}Dμ1y=Dμx is a one-to-one map from T U x T U x TU_(x)T U_{x}TUx to T U y T U y TU_(y)T U_{y}TUy. We will denote this map as D μ x y D μ x y Dmu_(xy)D \mu_{x y}Dμxy. We say a k k kkk-form, ω ω omega\omegaω, on R n R n R^(n)\mathrm{R}^{n}Rn descends to a k k kkk-form on U / μ 1 U / μ 1 U//mu^(-1)U / \mu^{-1}U/μ1 if and only if ω ( V 1 x , , V x x ) = ω ω V 1 x , , V x x = ω omega(V^(1)_(x),dots,V_(x)_(x))=omega\omega\left(V^{1}{ }_{x}, \ldots, V_{x}{ }_{x}\right)=\omegaω(V1x,,Vxx)=ω ( D μ x y ( V 1 ) , , D μ x y ( V 1 ) ) D μ x y V 1 , , D μ x y V 1 (Dmu_(xy)(V^(1)),dots,Dmu_(xy)(V^(1)))\left(D \mu_{x y}\left(V^{1}\right), \ldots, D \mu_{x y}\left(V^{1}\right)\right)(Dμxy(V1),,Dμxy(V1)), for all x , y U x , y U x,y in Ux, y \in Ux,yU such that μ ( x ) = μ ( y ) μ ( x ) = μ ( y ) mu(x)=mu(y)\mu(x)=\mu(y)μ(x)=μ(y).
9.12. If T is a differential k k kkk-form on M , then μ T M , then  μ T M_(", then ")mu_(***)TM_{\text {, then }} \mu_{\star} \mathrm{T}M, then μT (the lift of T ) is a differential k k kkk-form on U U UUU which descends to a differential k k kkk-form on U / μ 1 U / μ 1 U//mu^(-1)U / \mu^{-1}U/μ1.
Now suppose that we have a k k kkk-form, ω ω omega\omegaω, on U U UUU which descends to a k k kkk-form on U / μ 1 U / μ 1 U//mu^(-1)U / \mu^{-1}U/μ1, where μ : U R n M R m μ : U R n M R m mu:U subR^(n)rarr M subR^(m)\mu: U \subset \mathrm{R}^{n} \rightarrow M \subset \mathrm{R}^{m}μ:URnMRm is a covering map. How can we get a k k kkk-form on M M MMM ? As we have already remarked, μ μ mu\muμ : U / μ 1 M U / μ 1 M U//mu^(-1)rarr MU / \mu^{-1} \rightarrow MU/μ1M is a one-to-one map. Hence, we can use it to push forward the form, ω ω omega\omegaω. In other words, we can define a k k kkk-form on M M MMM as follows: Given k k kkk vectors in T M p T M p TM_(p)T M_{p}TMp, we first choose a point, x x x inx \inx μ 1 ( p ) μ 1 ( p ) mu^(-1)(p)\mu^{-1}(p)μ1(p). We then define
μ ω ( V p 1 , , V p k ) = ω ~ ( D μ x 1 ( V p 1 ) , , D μ x 1 ( V p k ) ) . μ ω V p 1 , , V p k = ω ~ D μ x 1 V p 1 , , D μ x 1 V p k . mu_(**)omega(V_(p)^(1),dots,V_(p)^(k))= tilde(omega)(Dmu_(x)^(-1)(V_(p)^(1)),dots,Dmu_(x)^(-1)(V_(p)^(k))).\mu_{*} \omega\left(V_{p}^{1}, \ldots, V_{p}^{k}\right)=\tilde{\omega}\left(D \mu_{x}^{-1}\left(V_{p}^{1}\right), \ldots, D \mu_{x}^{-1}\left(V_{p}^{k}\right)\right) .μω(Vp1,,Vpk)=ω~(Dμx1(Vp1),,Dμx1(Vpk)).
It follows from the fact that ω ω omega\omegaω descends to a form on U / μ 1 U / μ 1 U//mu^(-1)U / \mu^{-1}U/μ1 that it does not matter which point, x x xxx, we choose in μ 1 ( p ) μ 1 ( p ) mu^(-1)(p)\mu^{-1}(p)μ1(p). Note that although μ μ mu\muμ is not one-to-one, D μ x D μ x Dmu_(x)D \mu_{x}Dμx is, so D μ 1 x D μ 1 x Dmu^(-1)_(x)D \mu^{-1}{ }_{\mathrm{x}}Dμ1x makes sense.
If we begin with a form on U U UUU, there is a slightly more general construction of a form on a quotient of U U UUU, which does not require the use of a covering map. Let Γ Γ Gamma\GammaΓ be a group of transformations of U U UUU . We say Γ Γ Gamma\GammaΓ acts discretely if for each p U p U p in Up \in UpU, there exists an ∈> 0 ∈> 0 ∈>0\in>0∈>0 such that N ( p ) N ( p ) N_(in)(p)N_{\in}(p)N(p) does not contain γ ( p ) γ ( p ) gamma(p)\gamma(p)γ(p), for any non-identity element, γ Γ γ Γ gamma in Gamma\gamma \in \GammaγΓ. If Γ Γ Gamma\GammaΓ acts discretely, then we can form the quotient of U U UUU by Γ Γ Gamma\GammaΓ, denoted U Γ U Γ U GammaU \GammaUΓ, as follows: p q p q p∼qp \sim qpq if there exists γ Γ γ Γ gamma in Gamma\gamma \in \GammaγΓ such that y ( p ) = q y ( p ) = q y(p)=qy(p)=qy(p)=q. (The fact that Γ Γ Gamma\GammaΓ acts discretely is what guarantees a "nice" topology on U Γ U Γ U GammaU \GammaUΓ.)
Now, suppose ω ω omega\omegaω is a k k kkk-form on U U UUU. We say ω ω omega\omegaω descends to a k k kkk form, ω ω omega\omegaω, on U / Γ U / Γ U//GammaU / \GammaU/Γ, if and only if ω ( V 1 p , , V p k ) = ω ( D γ ( V p 1 ) , ω V 1 p , , V p k = ω D γ V p 1 , omega(V^(1)_(p),dots,V_(p)^(k))=omega(D_(gamma)(V_(p)^(1)),dots:}\omega\left(V^{1}{ }_{p}, \ldots, V_{p}^{k}\right)=\omega\left(D_{\gamma}\left(V_{p}^{1}\right), \ldots\right.ω(V1p,,Vpk)=ω(Dγ(Vp1),, Dy ( V 1 p ) ) Dy V 1 p {: Dy(V^(1)_(p)))\left.\operatorname{Dy}\left(V^{1}{ }_{p}\right)\right)Dy(V1p)), for all γ Γ γ Γ gamma in Gamma\gamma \in \GammaγΓ.
Now that we have decided what a form on a quotient of U U UUU is, we still have to define n n nnn-chains, and what we mean by integration of n n nnn forms over n n nnn-chains. We say an n n nnn-chain, C ~ U C ~ U tilde(C)sub U\tilde{C} \subset UC~U, descends to an n n nnn chain, C U / Γ C U / Γ C sub U//GammaC \subset U / \GammaCU/Γ, if y ( C ~ ) = C ~ y ( C ~ ) = C ~ y( tilde(C))= tilde(C)y(\tilde{C})=\tilde{C}y(C~)=C~, for all γ Γ γ Γ gamma in Gamma\gamma \in \GammaγΓ. The n n nnn-chains of U / Γ U / Γ U//GammaU / \GammaU/Γ are simply those which are descendants of n n nnn-chains in U U UUU.
Integration is a little more subtle. For this we need the concept of a fundamental domain for Γ Γ Gamma\GammaΓ. This is nothing more than a closed subset of U U UUU, whose interior does not contain two equivalent points. Furthermore, for each equivalence class, there is at least one representative in a fundamental domain. Here is one way to construct a fundamental domain: First, choose a point, p U p U p in Up \in UpU. Now, let D = { q U d ( p , q ) d ( γ ( p ) , q ) D = { q U d ( p , q ) d ( γ ( p ) , q ) D={q in U∣d(p,q) <= d(gamma(p),q)D=\{q \in U \mid d(p, q) \leq d(\gamma(p), q)D={qUd(p,q)d(γ(p),q), for all γ Γ } γ Γ } gamma in Gamma}\gamma \in \Gamma\}γΓ}.
Now, let C ~ C ~ tilde(C)\tilde{C}C~ be an n n nnn-chain on U U UUU which descends to an n n nnn-chain, C C CCC, on U / Γ U / Γ U//GammaU / \GammaU/Γ, and let ω ω omega\omegaω be an n n nnn-form that descends to an n n nnn-form, ω ω omega\omegaω. Let D D DDD be a fundamental domain for Γ Γ Gamma\GammaΓ in U U UUU. Then we define
C ω C ~ D ω ~ C ω C ~ D ω ~ int_(C)omega-=int_( tilde(C)nn D) tilde(omega)\int_{C} \omega \equiv \int_{\tilde{C} \cap D} \tilde{\omega}CωC~Dω~
Technical note: In general, this definition is invariant of which point was chosen in the construction of the fundamental domain, D D DDD. However, a VERY unlucky choice will result in C ~ D D C ~ D D tilde(C)nn D sub del D\tilde{C} \cap D \subset \partial DC~DD, which could give a different answer for the above integral. Fortunately, it can be shown that the set of such "unlucky" points has measure zero. That is, if we were to choose the point at random, then the odds of picking an "unlucky" point are 0 % 0 % 0%0 \%0%. Very unlucky indeed!
Example 46. Suppose Γ Γ Gamma\GammaΓ is the group of transformations of the plane generated by ( x , y ) ( x + 1 , y ) ( x , y ) ( x + 1 , y ) (x,y)rarr(x+1,y)(x, y) \rightarrow(x+1, y)(x,y)(x+1,y), and ( x , y ) ( x , y + 1 ) ( x , y ) ( x , y + 1 ) (x,y)rarr(x,y+1)(x, y) \rightarrow(x, y+1)(x,y)(x,y+1). The space R 2 / Γ R 2 / Γ R^(2)//Gamma\mathrm{R}^{2} / \GammaR2/Γ is often denoted T 2 T 2 T^(2)T^{2}T2, and referred to as a torus. Topologists often visualize the torus as the surface of a donut. A fundamental domain for Γ Γ Gamma\GammaΓ is the unit square in R 2 R 2 R^(2)\mathrm{R}^{2}R2. The 1 -form, d x d x dxd xdx, on R 2 R 2 R^(2)\mathrm{R}^{2}R2 descends to a 1 -form on T 2 T 2 T^(2)T^{2}T2. Integration of this form over a closed 1chain, C T 2 C T 2 C subT^(2)C \subset T^{2}CT2, counts the number of times C C CCC wraps around the "hole" of the donut.

9.5 Defining manifolds

As we have already remarked, a differential n n nnn-form on R m R m R^(m)\mathrm{R}^{m}Rm is just an n n nnn-form on T p R m T p R m T_(p)R^(m)T_{p} \mathrm{R}^{m}TpRm, for each point p R m p R m p inR^(m)p \in \mathrm{R}^{m}pRm, along with some condition about how the form varies in a differentiable way as p p ppp varies. All we need to define a form on a space other than R m R m R^(m)\mathrm{R}^{m}Rm is some notion of a tangent space at every point. We call such a space a manifold. In addition, we insist that at each point of a manifold the tangent space has the same dimension, n n nnn, which we then say is the dimension of the manifold.
How do we guarantee that a given subset of R m R m R^(m)\mathrm{R}^{m}Rm is a manifold? Recall that we defined the tangent space to be the span of some partial derivatives of a parameterization. However, insisting that the whole manifold is capable of being parameterized is very restrictive. Instead, we only insist that every point of a manifold lies in a subset that can be parameterized. Hence, if M M MMM is an n n nnn-manifold in R m R m R^(m)\mathrm{R}^{m}Rm then there is a set of open subsets, { U i } R n U i R n {U_(i)}subR^(n)\left\{U_{i}\right\} \subset \mathrm{R}^{n}{Ui}Rn, and a set of differentiable maps, { φ i : U i M } φ i : U i M {varphi_(i):U_(i)rarr M}\left\{\varphi_{i}: U_{i} \rightarrow M\right\}{φi:UiM}, such that { φ i ( U i ) } φ i U i {varphi_(i)(U_(i))}\left\{\varphi_{i}\left(U_{i}\right)\right\}{φi(Ui)} is a cover of M M MMM. (That is, for each point, p M p M p in Mp \in MpM, there is an i i iii, and a point, q U i q U i q inU_(i)q \in U_{i}qUi, such that φ i φ i varphi_(i)\varphi_{i}φi ( q q qqq ) = p = p =p=p=p.)
Example 47. S 1 S 1 S^(1)S^{1}S1, the unit circle in R 2 R 2 R^(2)R^{2}R2, is a 1-manifold. Let U i = ( 11 ) U i = ( 11 ) U_(i)=(-11)U_{i}=(-11)Ui=(11), for i = 1 , 2 , 3 , 4 , φ 1 ( t ) = ( t , 1 t 2 , φ 2 ( t ) = ( t , 1 t 2 ) , φ 3 ( t ) = ( 1 i = 1 , 2 , 3 , 4 , φ 1 ( t ) = t , 1 t 2 , φ 2 ( t ) = t , 1 t 2 , φ 3 ( t ) = ( 1 i=1,2,3,4,varphi_(1)(t)=(t,sqrt()1-t^(2),varphi_(2)(t)=(t,-sqrt()1-t^(2)),varphi_(3)(t)=(sqrt()1-:}i=1,2,3,4, \varphi_{1}(t)=\left(t, \sqrt{ } 1-\mathrm{t}^{2}, \varphi_{2}(t)=\left(t,-\sqrt{ } 1-\mathrm{t}^{2}\right), \varphi_{3}(t)=(\sqrt{ } 1-\right.i=1,2,3,4,φ1(t)=(t,1t2,φ2(t)=(t,1t2),φ3(t)=(1 t 2 , t ) t 2 , t {:t^(2),t)\left.\mathrm{t}^{2}, t\right)t2,t) and φ 4 ( t ) = ( 1 t 2 , t ) φ 4 ( t ) = 1 t 2 , t varphi_(4)(t)=(-sqrt(1-t^(2)),t)\varphi_{4}(t)=\left(-\sqrt{1-\mathrm{t}^{2}}, \mathrm{t}\right)φ4(t)=(1t2,t). Then { φ i ( U i ) } φ i U i {varphi_(i)(U_(i))}\left\{\varphi_{i}\left(U_{i}\right)\right\}{φi(Ui)} is certainly a cover of S 1 S 1 S^(1)S^{1}S1 with the desired properties.
9.13. Show that S 2 S 2 S^(2)S^{2}S2, the unit sphere in R 3 R 3 R^(3)\mathrm{R}^{3}R3, is a 2-manifold.

9.6 Differential forms on manifolds

Basically, the definition of a differential n n nnn-form on an m m mmm-manifold is the same as the definition of an n n nnn-form on a subset of R m R m R^(m)\mathrm{R}^{m}Rm which was given by a single parameterization. First and foremost it is just an n n nnn form on T p M T p M T_(p)MT_{p} MTpM, for each p M p M p in Mp \in MpM.
Let's say M M MMM is an m m mmm-manifold. Then we know there is a set of open sets, { U i } R m U i R m {U_(i)}subR^(m)\left\{U_{i}\right\} \subset R^{m}{Ui}Rm, and a set of differentiable maps, { φ i : U i M } φ i : U i M {varphi_(i):U_(i)rarr M}\left\{\varphi_{i}: U_{i} \rightarrow M\right\}{φi:UiM}, such that { Φ i ( U i ) } Φ i U i {Phi_(i)(U_(i))}\left\{\Phi_{i}\left(U_{i}\right)\right\}{Φi(Ui)} covers M M MMM. Now, let's say that ω ω omega\omegaω is a family of n n nnn-forms, defined on T p M T p M T_(p)MT_{p} MTpM, for each p M p M p in Mp \in MpM. Then we say that the family, ω ω omega\omegaω, is a differentiable n n nnn-form on M M MMM if φ , ω φ , ω varphi,omega\varphi, \omegaφ,ω is a differentiable n n nnn-form on U i , U i , U_(i,)U_{i,}Ui,, for each i i iii.
Example 48. In the previous section, we saw how S 1 S 1 S^(1)S^{1}S1, the unit circle in R 2 R 2 R^(2)\mathrm{R}^{2}R2, is a 1-manifold. If ( x , y ) ( x , y ) (x,y)(x, y)(x,y) is a point of S 1 S 1 S^(1)S^{1}S1, then T S 1 ( x , y ) T S 1 ( x , y ) TS^(1)(x,y)T S^{1}(x, y)TS1(x,y) is given by the equation d y = x / y d x d y = x / y d x dy=-x//ydxd y=-x / y d xdy=x/ydx, in RR 2 ( x , y ) RR 2 ( x , y ) RR^(2)(x,y)\operatorname{RR}^{2}(x, y)RR2(x,y), as long as y 0 y 0 y!=0y \neq 0y0. If y = 0 y = 0 y=0y=0y=0, then T S 1 ( x , y ) T S 1 ( x , y ) TS^(1)(x,y)T S^{1}(x, y)TS1(x,y) is given by d x = 0 d x = 0 dx=0d x=0dx=0. We define a 1 -form on S 1 , ω = y S 1 , ω = y S^(1),omega=-yS^{1}, \omega=-yS1,ω=y d x + x d y d x + x d y dx+xdyd x+x d ydx+xdy. (Actually, ω ω omega\omegaω is a 1 -form on all of R 2 R 2 R^(2)\mathrm{R}^{2}R2. To get a 1 -form on just S 1 S 1 S^(1)S^{1}S1, we restrict the domain of ω ω omega\omegaω to the tangent lines to S 1 S 1 S^(1)S^{1}S1.) To check that this is really a differential form, we must compute all pullbacks:
ϕ 1 ω = 1 1 t 2 d t , ϕ 2 ω = 1 1 t 2 d t ϕ 3 ω = 1 1 t 2 d t , ϕ 4 ω = 1 1 t 2 d t . ϕ 1 ω = 1 1 t 2 d t , ϕ 2 ω = 1 1 t 2 d t ϕ 3 ω = 1 1 t 2 d t , ϕ 4 ω = 1 1 t 2 d t . {:[phi_(1)^(**)omega=(-1)/(sqrt(1-t^(2)))dt","phi_(2)^(**)omega=(1)/(sqrt(1-t^(2)))dt],[phi_(3)^(**)omega=(1)/(sqrt(1-t^(2)))dt","phi_(4)^(**)omega=(-1)/(sqrt(1-t^(2)))dt.]:}\begin{aligned} & \phi_{1}^{*} \omega=\frac{-1}{\sqrt{1-t^{2}}} d t, \phi_{2}^{*} \omega=\frac{1}{\sqrt{1-t^{2}}} d t \\ & \phi_{3}^{*} \omega=\frac{1}{\sqrt{1-t^{2}}} d t, \phi_{4}^{*} \omega=\frac{-1}{\sqrt{1-t^{2}}} d t . \end{aligned}ϕ1ω=11t2dt,ϕ2ω=11t2dtϕ3ω=11t2dt,ϕ4ω=11t2dt.
Since all of these are differentiable on U i = ( 1 , 1 ) U i = ( 1 , 1 ) U_(i)=(-1,1)U_{i}=(-1,1)Ui=(1,1), we can say that ω ω omega\omegaω is a differential form on S 1 S 1 S^(1)S^{1}S1.
We now move on to integration of n n nnn-chains on manifolds. The definition of an n n nnn-chain is no different than before; it is just a formal linear combination of n n nnn-cells in M M MMM. Let's suppose that C C CCC is an n n nnn-chain in M M MMM, and ω ω omega\omegaω is an n n nnn-form. Then how do we definec ω ω omega\omegaω ? If C C CCC lies entirely in φ i ( U i ) φ i U i varphi_(i)(U_(i))\varphi_{i}\left(U_{i}\right)φi(Ui), for some i i iii, then we could define the value of this
integral to be the value of ϕ i 1 ( C ) ϕ i 1 ( C ) phi_(i)^(-1)(C)\phi_{i}^{-1}(C)ϕi1(C). But it may be that part of C C CCC lies in both φ i ( U i ) φ i U i varphi_(i)(U_(i))\varphi_{i}\left(U_{i}\right)φi(Ui) and φ j ( U j ) φ j U j varphi_(j)(U_(j))\varphi_{j}\left(U_{j}\right)φj(Uj). If we definec ω ω omega\omegaω to be the sum of the two integrals we get when we pull-back ω ω omega\omegaω under φ i φ i varphi_(i)\varphi_{i}φi and φ j φ j varphi_(j)\varphi_{j}φj, then we end up "double counting" the integral of ω ω omega\omegaω on C φ i ( U i ) φ j ( U j ) C φ i U i φ j U j C nnvarphi_(i)(U_(i))nnvarphi_(j)(U_(j))C \cap \varphi_{i}\left(U_{i}\right) \cap \varphi_{j}\left(U_{j}\right)Cφi(Ui)φj(Uj). Somehow, as we move from φ i ( U i ) φ i U i varphi_(i)(U_(i))\varphi_{i}\left(U_{i}\right)φi(Ui) into φ j ( U j ) φ j U j varphi_(j)(U_(j))\varphi_{j}\left(U_{j}\right)φj(Uj), we want the effect of the pull-back of ω ω omega\omegaω under φ i φ i varphi_(i)\varphi_{i}φi to "fade out," and the effect of the pull-back under φ j φ j varphi_(j)\varphi_{j}φj to "fade in." This is accomplished by a partition of unity.
The technical definition of a partition of unity subordinate to the cover, { φ i ( U i ) } φ i U i {varphi_(i)(U_(i))}\left\{\varphi_{i}\left(U_{i}\right)\right\}{φi(Ui)} is a set of differentiable functions, f i : M [ 0 , 1 ] f i : M [ 0 , 1 ] f_(i):M rarr[0,1]f_{i}: M \rightarrow[0,1]fi:M[0,1], such that f i ( p ) = 0 f i ( p ) = 0 f_(i)(p)=0f_{i}(p)=0fi(p)=0 if p φ i ( U i ) p φ i U i p!invarphi_(i)(U_(i))p \notin \varphi_{i}\left(U_{i}\right)pφi(Ui), and i f i ( p ) = 1 i f i ( p ) = 1 sum_(i)f_(i)(p)=1\sum_{i} f_{i}(p)=1ifi(p)=1, for all p M p M p in Mp \in MpM. We refer the reader to any book on differential topo gy for a proof of the existence of partitions of unity.
We are now ready to give the full definition of the integral of an n n nnn form on an n n nnn-chain in an m m mmm-manifold.
C ω i ϕ i 1 ( C ) ϕ i ( f i ω ) . C ω i ϕ i 1 ( C ) ϕ i f i ω . int_(C)omega-=sum_(i)int_(phi_(i)^(1)(C))phi_(i)^(**)(f_(i)omega).\int_{C} \omega \equiv \sum_{i} \int_{\phi_{i}^{1}(\mathcal{C})} \phi_{i}^{*}\left(f_{i} \omega\right) .Cωiϕi1(C)ϕi(fiω).
We illustrate this with a simple example.
Example 49. Let M M MMM be the manifold which is the interval ( 1 , 10 ) R ( 1 , 10 ) R (1,10)sub R(1,10) \subset R(1,10)R. Let U i = ( i , i + 2 ) U i = ( i , i + 2 ) U_(i)=(i,i+2)U_{i}=(i, i+2)Ui=(i,i+2), for i = 1 , , 8 i = 1 , , 8 i=1,dots,8i=1, \ldots, 8i=1,,8. Let φ i : U i M φ i : U i M varphi_(i):U_(i)rarr M\varphi_{i}: U_{i} \rightarrow Mφi:UiM be the identity map. Let { f i } f i {f_(i)}\left\{f_{i}\right\}{fi} be a partition of unity, subordinate to the cover, { φ { φ {varphi\{\varphi{φ i ( U i ) } i U i {:_(i)(U_(i))}\left.{ }_{i}\left(U_{i}\right)\right\}i(Ui)}. Let ω ω omega\omegaω be a 1-form on M. Finally, let C C CCC be the 1 -chain which consists of the single 1-cell, [ 2 , 8 ] [ 2 , 8 ] [2,8][2,8][2,8]. Then we have
C ω i = 1 8 ϕ i 1 ( C ) ϕ i ( f i ω ) = i = 1 8 C f i ω = C i = 1 8 ( f i ω ) = C ( i = 1 8 f i ) ω = C ω C ω i = 1 8 ϕ i 1 ( C ) ϕ i f i ω = i = 1 8 C f i ω = C i = 1 8 f i ω = C i = 1 8 f i ω = C ω int_(C)omega-=sum_(i=1)^(8)int_(phi_(i)^(-1)(C))phi_(i)^(**)(f_(i)omega)=sum_(i=1)^(8)int_(C)f_(i)omega=int_(C)sum_(i=1)^(8)(f_(i)omega)=int_(C)(sum_(i=1)^(8)f_(i))omega=int_(C)omega\int_{C} \omega \equiv \sum_{i=1}^{8} \int_{\phi_{i}^{-1}(C)} \phi_{i}^{*}\left(f_{i} \omega\right)=\sum_{i=1}^{8} \int_{C} f_{i} \omega=\int_{C} \sum_{i=1}^{8}\left(f_{i} \omega\right)=\int_{C}\left(\sum_{i=1}^{8} f_{i}\right) \omega=\int_{C} \omegaCωi=18ϕi1(C)ϕi(fiω)=i=18Cfiω=Ci=18(fiω)=C(i=18fi)ω=Cω
as one would expect!
Example 50. Let S 1 , U i j φ i S 1 , U i j φ i S^(1),U_(ij)varphi_(i)S^{1}, U_{i j} \varphi_{i}S1,Uijφi and ω ω omega\omegaω be defined as in Examples 47 and 48. A partition of unity subordinate to the cover { φ i ( U i ) } φ i U i {varphi_(i)(U_(i))}\left\{\varphi_{i}\left(U_{i}\right)\right\}{φi(Ui)} is as follows:
f 1 ( x , y ) = { y 2 y 0 0 y < 0 f 2 ( x , y ) = { 0 y > 0 y 2 y 0 f 3 ( x , y ) = { x 2 x 0 0 x < 0 f 4 ( x , y ) = { 0 x > 0 x 2 x 0 f 1 ( x , y ) = y 2 y 0 0 y < 0 f 2 ( x , y ) = 0 y > 0 y 2 y 0 f 3 ( x , y ) = x 2 x 0 0 x < 0 f 4 ( x , y ) = 0 x > 0 x 2 x 0 {:[f_(1)(x","y)={[y^(2),y >= 0],[0,y < 0]quadf_(2)(x,y)={[0,y > 0],[y^(2),y <= 0]:}],[f_(3)(x","y)={[x^(2),x >= 0],[0,x < 0]quadf_(4)(x,y)={[0,x > 0],[x^(2),x <= 0]:}]:}\begin{aligned} & f_{1}(x, y)=\left\{\begin{array}{lll} y^{2} & y \geq 0 \\ 0 & y<0 \end{array} \quad f_{2}(x, y)= \begin{cases}0 & y>0 \\ y^{2} & y \leq 0\end{cases} \right. \\ & f_{3}(x, y)=\left\{\begin{array}{ll} x^{2} & x \geq 0 \\ 0 & x<0 \end{array} \quad f_{4}(x, y)= \begin{cases}0 & x>0 \\ x^{2} & x \leq 0\end{cases} \right. \end{aligned}f1(x,y)={y2y00y<0f2(x,y)={0y>0y2y0f3(x,y)={x2x00x<0f4(x,y)={0x>0x2x0
(Check this!) Let μ : [ 0 , π ] S 1 μ : [ 0 , π ] S 1 mu:[0,pi]rarrS^(1)\mu:[0, \pi] \rightarrow S^{1}μ:[0,π]S1 be defined by μ ( θ ) = ( cos θ , sin θ μ ( θ ) = ( cos θ , sin θ mu(theta)=(cos theta,sin theta\mu(\theta)=(\cos \theta, \sin \thetaμ(θ)=(cosθ,sinθ ). Then the image of μ μ mu\muμ is a 1 -cell, σ σ sigma\sigmaσ, in S 1 S 1 S^(1)S^{1}S1. Let's integrate ω ω omega\omegaω over σ σ sigma\sigmaσ
σ ω i = 1 4 ϕ i 1 ( σ ) ϕ i ( f i ω ) = ( 1 , 1 ) 1 t 2 d t + 0 + [ 0 , 1 ) 1 t 2 d t + [ 0 , 1 ) 1 t 2 d t = 1 1 1 t 2 d t + 2 0 1 1 t 2 d t = π σ ω i = 1 4 ϕ i 1 ( σ ) ϕ i f i ω = ( 1 , 1 ) 1 t 2 d t + 0 + [ 0 , 1 ) 1 t 2 d t + [ 0 , 1 ) 1 t 2 d t = 1 1 1 t 2 d t + 2 0 1 1 t 2 d t = π {:[int_(sigma)omega-=sum_(i=1)^(4)int_(phi_(i)^(-1)(sigma))phi_(i)^(**)(f_(i)omega)],[=int_(-(-1,1))-sqrt(1-t^(2))dt+0+int_([0,1))sqrt(1-t^(2))dt+int_(-[0,1))-sqrt(1-t^(2))dt],[=int_(-1)^(1)sqrt(1-t^(2))dt+2int_(0)^(1)sqrt(1-t^(2))dt],[=pi]:}\begin{aligned} \int_{\sigma} \omega & \equiv \sum_{i=1}^{4} \int_{\phi_{i}^{-1}(\sigma)} \phi_{i}^{*}\left(f_{i} \omega\right) \\ & =\int_{-(-1,1)}-\sqrt{1-t^{2}} d t+0+\int_{[0,1)} \sqrt{1-t^{2}} d t+\int_{-[0,1)}-\sqrt{1-t^{2}} d t \\ & =\int_{-1}^{1} \sqrt{1-t^{2}} d t+2 \int_{0}^{1} \sqrt{1-t^{2}} d t \\ & =\pi \end{aligned}σωi=14ϕi1(σ)ϕi(fiω)=(1,1)1t2dt+0+[0,1)1t2dt+[0,1)1t2dt=111t2dt+2011t2dt=π
CAUTION: Beware of orientations!

9.7 Application: DeRham cohomology

One of the predominant uses of differential forms is to give global information about manifolds. Consider the space R 2 ( 0 , 0 ) R 2 ( 0 , 0 ) R^(2)-(0,0)R^{2}-(0,0)R2(0,0), as in Example 44. Near every point of this space we can find an open set which is identical to an open set around a point of R 2 R 2 R^(2)R^{2}R2. This means that all of the local information in R 2 ( 0 , 0 ) R 2 ( 0 , 0 ) R^(2)-(0,0)R^{2}-(0,0)R2(0,0) is the same as the local information in R 2 R 2 R^(2)R^{2}R2. The fact that the origin is missing is a global property.
For the purposes of detecting global properties, certain forms are interesting, and certain forms are completely uninteresting. We will spend some time discussing both. The interesting forms are the ones whose derivative is zero. Such forms are said to be closed. An example of a closed 1 -form was ω 0 ω 0 omega_(0)\omega_{0}ω0, from Example 44 of the previous chapter. For now, let's just focus on closed 1 -forms so that you can keep this example in mind.
Let's look at what happens when we integrate a closed 1-form ω 0 ω 0 omega_(0)\omega_{0}ω0 over a 1 -chain C C CCC such that C = 0 C = 0 del C=0\partial C=0C=0 (i.e., C C CCC is a closed 1 -chain). If C C CCC bounds a disk D D DDD then Stokes' theorem says
C ω 0 = D d ω 0 = D 0 = 0 . C ω 0 = D d ω 0 = D 0 = 0 . int_(C)omega_(0)=int_(D)domega_(0)=int_(D)0=0.\int_{C} \omega_{0}=\int_{D} d \omega_{0}=\int_{D} 0=0 .Cω0=Ddω0=D0=0.
In a sufficiently small region of every manifold, every closed 1-chain bounds a disk. So integrating closed 1 -forms on "small" 1 -chains gives us no information. In other words, closed 1 -forms give no local information.
Suppose now that we have a closed 1-form ω 0 ω 0 omega_(0)\omega_{0}ω0 and a closed 1chain C C CCC such that C ω 0 0 C ω 0 0 int_(C)omega_(0)!=0\int_{C} \omega_{0} \neq 0Cω00. Then we know C C CCC does not bound a disk. The fact that there exists such a 1 -chain is global information. This is why we say that the closed forms are the ones that are interesting, from the point of view of detecting only global information.
Now let's suppose that we have a 1 -form ω 1 ω 1 omega_(1)\omega_{1}ω1, that is the derivative of a 0 -form f f fff (i.e., ω 1 = d f ω 1 = d f omega_(1)=df\omega_{1}=d fω1=df ). We say such a form is exact. Again, let C C CCC be a closed 1 -chain. Let's pick two points, p p ppp and q q qqq, on C C CCC. Then C = C = C=C=C= C 1 + C 2 C 1 + C 2 C_(1)+C_(2)C_{1}+C_{2}C1+C2, where C 1 C 1 C_(1)C_{1}C1 goes from p p ppp to q q qqq and C 2 C 2 C_(2)C_{2}C2 goes from q q qqq back to p p ppp. Now let's do a quick computation:
C ω 1 = C 1 + C 2 ω 1 = C 1 ω 1 + C 2 ω 1 = C 1 d f + C 2 d f = p q f + q p f = 0 . C ω 1 = C 1 + C 2 ω 1 = C 1 ω 1 + C 2 ω 1 = C 1 d f + C 2 d f = p q f + q p f = 0 . {:[int_(C)omega_(1)=int_(C_(1)+C_(2))omega_(1)],[=int_(C_(1))omega_(1)+int_(C_(2))omega_(1)],[=int_(C_(1))df+int_(C_(2))df],[=int_(p-q)f+int_(q-p)f],[=0.]:}\begin{aligned} \int_{C} \omega_{1} & =\int_{C_{1}+C_{2}} \omega_{1} \\ & =\int_{C_{1}} \omega_{1}+\int_{C_{2}} \omega_{1} \\ & =\int_{C_{1}} d f+\int_{C_{2}} d f \\ & =\int_{p-q} f+\int_{q-p} f \\ & =0 . \end{aligned}Cω1=C1+C2ω1=C1ω1+C2ω1=C1df+C2df=pqf+qpf=0.
So integrating an exact form over a closed 1-chain always gives zero. This is why we say the exact forms are completely uninteresting. Unfortunately, in Problem 6.9 we learned that every exact form is also closed. This is a problem, since this would say that all of the completely uninteresting forms are also interesting! To remedy this we define an equivalence relation.
We pause here for a moment to explain what this means. An equivalence relation is just a way of taking one set and creating a new set by declaring certain objects in the original set to be "the same." This is the idea behind telling time. To construct the clock numbers, start with the integers and declare two to be "the same" if they differ by a multiple of 12 . So 10 + 3 = 13 10 + 3 = 13 10+3=1310+3=1310+3=13, but 13 is the same
as one, so if it's now 10 o'clock then in three hours it will one o'clock.
We play the same trick for differential forms. We will restrict ourselves to the closed forms, but we will consider two of them to be "the same" if their difference is an exact form. The set which we end up with is called the cohomology of the manifold in question. For example, if we start with the closed 1 -forms, then, after our equivalence relation, we end up with the set which we will call H 1 H 1 H^(1)H^{1}H1, or the first cohomology (see Figure 9.2).
Fig. 9.2. Defining H n H n H^(n)H^{n}Hn.
Note that the difference between an exact form and the form which always returns the number zero is an exact form. Hence, every exact form is equivalent to 0 in H n H n H^(n)H^{n}Hn, as in the figure.
For each n n nnn the set H n H n H^(n)H^{n}Hn contains a lot of information about the manifold in question. For example, if H 1 R 1 H 1 R 1 H^(1)~=R^(1)H^{1} \cong R^{1}H1R1 (as it turns out is the case for R 2 ( 0 , 0 ) R 2 ( 0 , 0 ) R^(2)-(0,0)R^{2}-(0,0)R2(0,0) ), then this tells us that the manifold has one "hole" in it. Studying manifolds via cohomology is one topic of a field of mathematics called Algebraic Topology. For a complete treatment of this subject, see [BT95].
A
Non-linear forms

A. 1 Surface area

Now that we have developed some proficiency with differential forms, let's see what else we can integrate. A basic assumption that we used to come up with the definition of an n n nnn-form was the fact that at every point it is a linear function which "eats" n n nnn vectors and returns a number. But what about the non-linear functions?
Let's go all the way back to Section 3.5. There we decided that the integral of a function f f fff over a surface R R RRR in R 3 R 3 R^(3)\mathrm{R}^{3}R3 should look something like:
(A.1) R f ( ϕ ( r , θ ) ) Area [ ϕ r ( r , θ ) , ϕ θ ( r , θ ) ] d r d θ . (A.1) R f ( ϕ ( r , θ ) ) Area ϕ r ( r , θ ) , ϕ θ ( r , θ ) d r d θ {:(A.1)int_(R)f(phi(r","theta))Area[(del phi)/(del r)(r,theta),(del phi)/(del theta)(r,theta)]drd theta". ":}\begin{equation*} \int_{R} f(\phi(r, \theta)) \operatorname{Area}\left[\frac{\partial \phi}{\partial r}(r, \theta), \frac{\partial \phi}{\partial \theta}(r, \theta)\right] d r d \theta \text {. } \tag{A.1} \end{equation*}(A.1)Rf(ϕ(r,θ))Area[ϕr(r,θ),ϕθ(r,θ)]drdθ
At the heart of the integrand is the Area function, which takes two vectors and returns the area of the parallelogram that it spans. The 2-form d x d y d x d y dx^^dyd x \wedge d ydxdy does this for two vectors in T p R 2 T p R 2 T_(p)R^(2)T_{p} \mathrm{R}^{2}TpR2. In T p R 3 T p R 3 T_(p)R^(3)T_{p} \mathrm{R}^{3}TpR3 the right function is the following:
Area ( V p 1 , V p 2 ) = ( d y d z ) 2 + ( d x d z ) 2 + ( d x d y ) 2 .  Area  V p 1 , V p 2 = ( d y d z ) 2 + ( d x d z ) 2 + ( d x d y ) 2 . " Area "(V_(p)^(1),V_(p)^(2))=sqrt((dy^^dz)^(2)+(dx^^dz)^(2)+(dx^^dy)^(2)).\text { Area }\left(V_{p}^{1}, V_{p}^{2}\right)=\sqrt{(d y \wedge d z)^{2}+(d x \wedge d z)^{2}+(d x \wedge d y)^{2}} . Area (Vp1,Vp2)=(dydz)2+(dxdz)2+(dxdy)2.
(The reader may recogni e this as the magnitude of the cross product between
V p 1 V p 1 V_(p)^(1)V_{p}^{1}Vp1 and V p 2 V p 2 V_(p)^(2)V_{p}^{2}Vp2.) This is clearly non-linear!
Example 51. The area of the parallelogram spanned by 1 , 1 , 0 1 , 1 , 0 (:1,1,0:)\langle 1,1,0\rangle1,1,0 and 1 , 2 , 3 1 , 2 , 3 (:1,2,3:)\langle 1,2,3\rangle1,2,3 can be computed as follows:
Area ( ( 1 , 1 , 0 , 1 , 2 , 3 ) = | 1 0 2 3 | 2 + | 1 0 | 2 + | 1 1 1 3 | 2 1 2 | 2 , ( 3 2 + 3 2 + 1 2 . Area ( ( 1 , 1 , 0 , 1 , 2 , 3 ) = 1 0 2 3 2 + 1 0 2 + 1 1 1 3 2 1 2 2 , 3 2 + 3 2 + 1 2 . [Area((1","1","0:)","(:1","2","3:)),=sqrt(|[1,0],[2,3]|^(2)+|[1,0]|^(2)+|[1,1],[1,3]|^(2))],[1,2]|^(2),(sqrt(3^(2)+3^(2)+1^(2)).:}\left.\begin{array}{rl} \operatorname{Area}((1,1,0\rangle,\langle 1,2,3\rangle) & =\sqrt{\left|\begin{array}{ll} 1 & 0 \\ 2 & 3 \end{array}\right|^{2}+\left|\begin{array}{ll} 1 & 0 \end{array}\right|^{2}+\left|\begin{array}{ll} 1 & 1 \\ 1 & 3 \end{array}\right|^{2}} \\ 1 & 2 \end{array}\right|^{2}, ~\left(\sqrt{3^{2}+3^{2}+1^{2}} .\right.Area((1,1,0,1,2,3)=|1023|2+|10|2+|1113|212|2, (32+32+12.
The thing that makes (linear) differential forms so useful is the generalized Stokes' Theorem. We do not have anything like this for non-linear forms, but that is not to say that they do not have their uses. For example, there is no differential 2-form on R 3 R 3 R^(3)\mathrm{R}^{3}R3 that one can integrate over arbitrary surfaces to find their surface area. For that we would need to compute the following:
Area ( R ) = S ( d y d z ) 2 + ( d x d z ) 2 + ( d x d y ) 2 Area ( R ) = S ( d y d z ) 2 + ( d x d z ) 2 + ( d x d y ) 2 Area(R)=int_(S)sqrt((dy^^dz)^(2)+(dx^^dz)^(2)+(dx^^dy)^(2))\operatorname{Area}(R)=\int_{S} \sqrt{(d y \wedge d z)^{2}+(d x \wedge d z)^{2}+(d x \wedge d y)^{2}}Area(R)=S(dydz)2+(dxdz)2+(dxdy)2
For relatively simple surfaces, this integrand can be evaluated by hand. Integrals such as this play a particularly important role in certain applied problems. For example, if one were to dip a loop of bent wire into a soap film, the resulting surface would be the one of minimal area. Before one can even begin to figure out what surface this is for a given piece of wire, one must be able to know how to compute the area of an arbitrary surface, as above.
Example 52. We compute the surface area of a sphere of radius r r rrr in R 3 R 3 R^(3)R^{3}R3. A parameterization is given by where 0 θ 2 π 0 θ 2 π 0 <= theta <= 2pi0 \leq \theta \leq 2 \pi0θ2π and 0 φ π 0 φ π 0 <= varphi <= pi0 \leq \varphi \leq \pi0φπ. Now we compute:
Φ ( θ , ϕ ) = ( r sin ϕ cos θ , r sin ϕ sin θ , r cos ϕ ) Φ ( θ , ϕ ) = ( r sin ϕ cos θ , r sin ϕ sin θ , r cos ϕ ) Phi(theta,phi)=(r sin phi cos theta,r sin phi sin theta,r cos phi)\Phi(\theta, \phi)=(r \sin \phi \cos \theta, r \sin \phi \sin \theta, r \cos \phi)Φ(θ,ϕ)=(rsinϕcosθ,rsinϕsinθ,rcosϕ)
Area ( ϕ θ , ϕ ϕ ) = Area ( r sin ϕ sin θ , r sin ϕ cos θ , 0 , r cos ϕ cos θ , r cos ϕ sin θ , r sin ϕ ) = ( r 2 sin 2 ϕ cos θ ) 2 + ( r 2 sin 2 ϕ sin θ ) 2 + ( r 2 sin ϕ cos ϕ ) 2 = r sin 4 ϕ + sin 2 ϕ cos 2 ϕ = r sin ϕ  Area  ϕ θ , ϕ ϕ =  Area  ( r sin ϕ sin θ , r sin ϕ cos θ , 0 , r cos ϕ cos θ , r cos ϕ sin θ , r sin ϕ ) = r 2 sin 2 ϕ cos θ 2 + r 2 sin 2 ϕ sin θ 2 + r 2 sin ϕ cos ϕ 2 = r sin 4 ϕ + sin 2 ϕ cos 2 ϕ = r sin ϕ {:[" Area "((del phi)/(del theta),(del phi)/(del phi))],[=" Area "((:-r sin phi sin theta","r sin phi cos theta","0:)","(:r cos phi cos theta","r cos phi sin theta","-r sin phi:))],[=sqrt((-r^(2)sin^(2)phi cos theta)^(2)+(r^(2)sin^(2)phi sin theta)^(2)+(-r^(2)sin phi cos phi)^(2))],[=rsqrt(sin^(4)phi+sin^(2)phicos^(2)phi)],[=r sin phi]:}\begin{aligned} & \text { Area }\left(\frac{\partial \phi}{\partial \theta}, \frac{\partial \phi}{\partial \phi}\right) \\ & =\text { Area }(\langle-r \sin \phi \sin \theta, r \sin \phi \cos \theta, 0\rangle,\langle r \cos \phi \cos \theta, r \cos \phi \sin \theta,-r \sin \phi\rangle) \\ & =\sqrt{\left(-r^{2} \sin ^{2} \phi \cos \theta\right)^{2}+\left(r^{2} \sin ^{2} \phi \sin \theta\right)^{2}+\left(-r^{2} \sin \phi \cos \phi\right)^{2}} \\ & =r \sqrt{\sin ^{4} \phi+\sin ^{2} \phi \cos ^{2} \phi} \\ & =r \sin \phi \end{aligned} Area (ϕθ,ϕϕ)= Area (rsinϕsinθ,rsinϕcosθ,0,rcosϕcosθ,rcosϕsinθ,rsinϕ)=(r2sin2ϕcosθ)2+(r2sin2ϕsinθ)2+(r2sinϕcosϕ)2=rsin4ϕ+sin2ϕcos2ϕ=rsinϕ
And so the desired area is given by
S Area ( Φ θ , Φ ϕ ) d θ d ϕ = 0 π 0 2 π r sin ϕ d θ d ϕ = 4 π r S Area Φ θ , Φ ϕ d θ d ϕ = 0 π 0 2 π r sin ϕ d θ d ϕ = 4 π r {:[int_(S)Area((del Phi)/(del theta),(del Phi)/(del phi))d theta d phi],[=int_(0)^(pi)int_(0)^(2pi)r sin phi d theta d phi],[=4pi r]:}\begin{aligned} & \int_{S} \operatorname{Area}\left(\frac{\partial \Phi}{\partial \theta}, \frac{\partial \Phi}{\partial \phi}\right) d \theta d \phi \\ = & \int_{0}^{\pi} \int_{0}^{2 \pi} r \sin \phi d \theta d \phi \\ = & 4 \pi r \end{aligned}SArea(Φθ,Φϕ)dθdϕ=0π02πrsinϕdθdϕ=4πr
A.1. Compute the surface area of a sphere of radius r r rrr in R 3 R 3 R^(3)\mathrm{R}^{3}R3 using the parameterizations for the top and bottom halves, where 0 ρ 0 ρ 0 <= rho <=0 \leq \rho \leq0ρ r a 0 θ 2 π r a 0 θ 2 π ra0 <= theta <= 2pir a 0 \leq \theta \leq 2 \pira0θ2π.
Φ ( ρ , θ ) = ( ρ cos θ , ρ sin θ , ± r 2 ρ 2 ) Φ ( ρ , θ ) = ρ cos θ , ρ sin θ , ± r 2 ρ 2 Phi(rho,theta)=(rho cos theta,rho sin theta,+-sqrt(r^(2)-rho^(2)))\Phi(\rho, \theta)=\left(\rho \cos \theta, \rho \sin \theta, \pm \sqrt{r^{2}-\rho^{2}}\right)Φ(ρ,θ)=(ρcosθ,ρsinθ,±r2ρ2)
Let's now go back to Equation A.1. Classically, this is called a surface integral. It might be a little clearer how to compute such an integral if we write it as follows:
R f ( x , y , z ) d S = R f ( x , y , z ) ( d y d z ) 2 + ( d x d z ) 2 + ( d x d y ) 2 R f ( x , y , z ) d S = R f ( x , y , z ) ( d y d z ) 2 + ( d x d z ) 2 + ( d x d y ) 2 int_(R)f(x,y,z)dS=int_(R)f(x,y,z)sqrt((dy^^dz)^(2)+(dx^^dz)^(2)+(dx^^dy)^(2))\int_{R} f(x, y, z) d S=\int_{R} f(x, y, z) \sqrt{(d y \wedge d z)^{2}+(d x \wedge d z)^{2}+(d x \wedge d y)^{2}}Rf(x,y,z)dS=Rf(x,y,z)(dydz)2+(dxdz)2+(dxdy)2

A. 2 Arc length

Lengths are very similar to areas. In calculus you learn that if you have a curve C C CCC in the plane, for example, parameterized by the function φ ( t ) = ( x ( t ) , y ( t ) ) φ ( t ) = ( x ( t ) , y ( t ) ) varphi(t)=(x(t),y(t))\varphi(t)=(x(t), y(t))φ(t)=(x(t),y(t)), where a t b a t b a <= t <= ba \leq t \leq batb, then its arc length is given by
Length ( C ) = a b ( d x d t ) 2 + ( d y d t ) 2 d t Length ( C ) = a b d x d t 2 + d y d t 2 d t Length(C)=int_(a)^(b)sqrt(((dx)/(dt))^(2)+((dy)/(dt))^(2))dt\operatorname{Length}(C)=\int_{a}^{b} \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d tLength(C)=ab(dxdt)2+(dydt)2dt
We can write this without making reference to the parameterization by employing a non-linear 1-form:
Length ( C ) = C d x 2 + d y 2  Length  ( C ) = C d x 2 + d y 2 " Length "(C)=int_(C)sqrt(dx^(2)+dy^(2))\text { Length }(C)=\int_{C} \sqrt{d x^{2}+d y^{2}} Length (C)=Cdx2+dy2
Finally, we can define what is classically called a line integral as follows:
C f ( x , y ) d s = C f ( x , y ) d x 2 + d y 2 . C f ( x , y ) d s = C f ( x , y ) d x 2 + d y 2 . oint_(C)f(x,y)ds=int_(C)f(x,y)sqrt(dx^(2)+dy^(2)).\oint_{C} f(x, y) d s=\int_{C} f(x, y) \sqrt{d x^{2}+d y^{2}} .Cf(x,y)ds=Cf(x,y)dx2+dy2.

References

[Arn97] V. I. Arnold. Mathematical Methods of Classical Mechanics. Springer, 1997.
[BT95] Raoul Bott and Loring Tu. Differential Forms in Algebraic Topology. Springer, 1995.
[GP74] Victor Guillemin and Alan Pollack. Differential Topology. Prentice Hall, 1974.
[HH01] John Hubbard and Barbara Hubbard. Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach. Prentice Hall, 2001.
[MTW73] Charles Misner, Kip Thorne, and John Wheeler. Gravitation. W. H. Freeman, 1973.

Index

n n nnn-form
0 -form
1-form
2-form
4-current
algebraic topology arc length
area form
boundary
cell
chain
change of variables formula
charge density closed chain
closed form
contact structure
cover
covering map
critical point
cross product
curl
current density
cylindrical coordinates
DeRham cohomology derivative of differential form of parameterized curve of parameterized surface partial
Descartes, René determinant
differential form
dimension
directional derivative
discrete action
div
dot product

equivalence relation exact form

Faraday

foliation
Fubini's theorem
fundamental domain
Fundamental Theorem of Calculus
Gauss' Divergence Theorem
Gobillion-Vey invariant
grad
gradient
gradient field
Green's theorem
group

Abstract

invariant

kernel
lattice
Legendrian curve level curve
lift
line field
line integral
linear function
manifold
Maxwell
Maxwell's equations
measure zero
multiple integral
open set orientation induced

parameterizationparameterizedcurvelineregionsurfacepartial derivativepartition of unityplane fieldpull-back

rectangluar coordinates
Reeb foliation
Riemann sum
  • n n nnn

second partial

spherical coordinates

Stokes' theorem classical
generalized
substitution rule
surface area
surface integral
tangent plane tangent space torus
transformation

vector

addition
unit
vector calculus
vector field
volume form
wedge product winding number

Solutions

Chapter 1

1.2 2 3 5 , 4 5 2 3 5 , 4 5 {:2^((:(3)/(5):}),(4)/(5):)\left.2^{\left\langle\frac{3}{5}\right.}, \frac{4}{5}\right\rangle235,45
  1. 4 8 5 15 4 8 5 15 4(8sqrt5)/(15)4 \frac{8 \sqrt{5}}{15}48515
    1.6 -8
    1.8
  2. x + y x + y x+yx+yx+y
  3. 4 x 2 y 2 4 x 2 y 2 sqrt(4-x^(2)-y^(2))\sqrt{4-x^{2}-y^{2}}4x2y2
  4. 1 ( 2 x 2 y 2 ) 2 1 2 x 2 y 2 2 sqrt(1-(2-x^(2)-y^(2))^(2))\sqrt{1-\left(2-x^{2}-y^{2}\right)^{2}}1(2x2y2)2
  5. 1 ( y x ) 2 1 ( y x ) 2 sqrt(1-(y-x)^(2))\sqrt{1-(y-x)^{2}}1(yx)2
    1.9
  6. 21
  7. 1
  8. 3
  9. 16
  10. 2 5 ( 33 2 5 3 5 ) 2 5 33 2 5 3 5 (2)/(5)(33-sqrt2^(5)-sqrt3^(5))\frac{2}{5}\left(33-\sqrt{2}^{5}-\sqrt{3}^{5}\right)25(332535)
    1.10
  11. f x = 2 x y 3 , f y = 3 x 2 y 2 f x = 2 x y 3 , f y = 3 x 2 y 2 (del f)/(del x)=2xy^(3),(del f)/(del y)=3x^(2)y^(2)\frac{\partial f}{\partial x}=2 x y^{3}, \frac{\partial f}{\partial y}=3 x^{2} y^{2}fx=2xy3,fy=3x2y2
  12. f x = 2 x y 3 cos ( x 2 y 3 ) , f y = 3 x 2 y 2 cos ( x 2 y 3 ) f x = 2 x y 3 cos x 2 y 3 , f y = 3 x 2 y 2 cos x 2 y 3 (del f)/(del x)=2xy^(3)cos(x^(2)y^(3)),(del f)/(del y)=3x^(2)y^(2)cos(x^(2)y^(3))\frac{\partial f}{\partial x}=2 x y^{3} \cos \left(x^{2} y^{3}\right), \frac{\partial f}{\partial y}=3 x^{2} y^{2} \cos \left(x^{2} y^{3}\right)fx=2xy3cos(x2y3),fy=3x2y2cos(x2y3)
  13. f x = sin ( x y ) + x y cos ( x y ) , f y = x 2 cos ( x y ) f x = sin ( x y ) + x y cos ( x y ) , f y = x 2 cos ( x y ) (del f)/(del x)=sin(xy)+xy cos(xy),(del f)/(del y)=x^(2)cos(xy)\frac{\partial f}{\partial x}=\sin (x y)+x y \cos (x y), \frac{\partial f}{\partial y}=x^{2} \cos (x y)fx=sin(xy)+xycos(xy),fy=x2cos(xy)
    1.12 4 2 1.12 4 2 1.12-4sqrt21.12-4 \sqrt{2}1.1242
    1.13 2 5 5 1.13 2 5 5 1.13^((-2sqrt5)/(5))1.13^{\frac{-2 \sqrt{5}}{5}}1.13255
    1.14
  14. y 2 , 2 x y y 2 , 2 x y (:y^(2),2xy:)\left\langle y^{2}, 2 x y\right\rangley2,2xy
  15. 69
  16. 9 15 , 12 15 9 15 , 12 15 (:(9)/(15),(12)/(15):)\left\langle\frac{9}{15}, \frac{12}{15}\right\rangle915,1215
  17. 15

Chapter 2

2.1

  1. b/a
  2. They are parallel. The one parameterized by φ φ varphi\varphiφ can be obtained from the other by shifting c c ccc units to the right and d d ddd units up.
    2.3
  3. ( cos 2 θ , cos θ sin θ ) cos 2 θ , cos θ sin θ (cos^(2)theta,cos theta sin theta)\left(\cos ^{2} \theta, \cos \theta \sin \theta\right)(cos2θ,cosθsinθ)
  4. ( x , sin x ) ( x , sin x ) (x,sin x)(x, \sin x)(x,sinx)
    2.4 φ ( t ) = ( t , 4 t 3 ) , 1 t 2 φ ( t ) = ( t , 4 t 3 ) , 1 t 2 varphi(t)=(t,4t-3),1 <= t <= 2\varphi(t)=(t, 4 t-3), 1 \leq t \leq 2φ(t)=(t,4t3),1t2 (There are many other answers.) 2.6
1. ( t 2 , t ) 2. 4 , 1  1.  t 2 , t  2.  4 , 1 {:[" 1. "(t^(2),t)],[" 2. "(:4","1:)]:}\begin{aligned} & \text { 1. }\left(t^{2}, t\right) \\ & \text { 2. }\langle 4,1\rangle \end{aligned} 1. (t2,t) 2. 4,1

2.7

r = ρ sin ϕ ρ = r 2 + z 2 θ = θ θ = θ z = ρ cos ϕ ϕ = tan 1 ( r z ) r = ρ sin ϕ ρ = r 2 + z 2 θ = θ θ = θ z = ρ cos ϕ ϕ = tan 1 r z {:[r=rho sin phi∣rho=sqrt(r^(2)+z^(2))],[theta=thetaquad theta=theta],[z=rho cos phi phi=tan^(-1)((r)/(z))]:}\begin{aligned} & r=\rho \sin \phi \mid \rho=\sqrt{r^{2}+z^{2}} \\ & \theta=\theta \quad \theta=\theta \\ & z=\rho \cos \phi \phi=\tan ^{-1}\left(\frac{r}{z}\right) \end{aligned}r=ρsinϕρ=r2+z2θ=θθ=θz=ρcosϕϕ=tan1(rz)
  1. z = x 2 + y 2 , z = r , ϕ = π 4 z = x 2 + y 2 , z = r , ϕ = π 4 z=sqrt(x^(2)+y^(2)),z=r,phi=(pi)/(4)z=\sqrt{x^{2}+y^{2}}, z=r, \phi=\frac{\pi}{4}z=x2+y2,z=r,ϕ=π4
  2. y = 0 , θ = 0 , θ = 0 y = 0 , θ = 0 , θ = 0 y=0,theta=0,theta=0y=0, \theta=0, \theta=0y=0,θ=0,θ=0
  3. z = 0 , z = 0 , ϕ = π 2 z = 0 , z = 0 , ϕ = π 2 z=0,z=0,phi=(pi)/(2)z=0, z=0, \phi=\frac{\pi}{2}z=0,z=0,ϕ=π2
  4. z = x + y , z = r ( sin θ + cos θ ) , cot ϕ = sin θ + cos θ z = x + y , z = r ( sin θ + cos θ ) , cot ϕ = sin θ + cos θ z=x+y,z=r(sin theta+cos theta),cot phi=sin theta+cos thetaz=x+y, z=r(\sin \theta+\cos \theta), \cot \phi=\sin \theta+\cos \thetaz=x+y,z=r(sinθ+cosθ),cotϕ=sinθ+cosθ
  5. z = ( x 2 + y 2 ) 3 , z = r 3 , z = ( ρ sin ϕ ) 3 z = x 2 + y 2 3 , z = r 3 , z = ( ρ sin ϕ ) 3 z=(x^(2)+y^(2))^(3),z=r^(3),z=(rho sin phi)^(3)z=\left(x^{2}+y^{2}\right)^{3}, z=r^{3}, z=(\rho \sin \phi)^{3}z=(x2+y2)3,z=r3,z=(ρsinϕ)3

2.10

  1. ϕ ( u , z ) = ( u , u , z ) ϕ ( u , z ) = ( u , u , z ) phi(u,z)=(u,u,z)\phi(u, z)=(u, u, z)ϕ(u,z)=(u,u,z)
  2. ϕ ( r , θ ) = ( r cos θ , r sin θ , r 2 ) ϕ ( r , θ ) = r cos θ , r sin θ , r 2 phi(r,theta)=(r cos theta,r sin theta,r^(2))\phi(r, \theta)=\left(r \cos \theta, r \sin \theta, r^{2}\right)ϕ(r,θ)=(rcosθ,rsinθ,r2)
  3. ψ ( θ , ϕ ) = ( ϕ sin ϕ cos θ , ϕ sin ϕ sin θ , ϕ cos ϕ ) ψ ( θ , ϕ ) = ( ϕ sin ϕ cos θ , ϕ sin ϕ sin θ , ϕ cos ϕ ) psi(theta,phi)=(phi sin phi cos theta,phi sin phi sin theta,phi cos phi)\psi(\theta, \phi)=(\phi \sin \phi \cos \theta, \phi \sin \phi \sin \theta, \phi \cos \phi)ψ(θ,ϕ)=(ϕsinϕcosθ,ϕsinϕsinθ,ϕcosϕ)
  4. ψ ( θ , ϕ ) = ( cos ϕ sin ϕ cos θ , cos ϕ sin ϕ sin θ , cos ϕ cos ϕ ) ψ ( θ , ϕ ) = ( cos ϕ sin ϕ cos θ , cos ϕ sin ϕ sin θ , cos ϕ cos ϕ ) psi(theta,phi)=(cos phi sin phi cos theta,cos phi sin phi sin theta,cos phi cos phi)\psi(\theta, \phi)=(\cos \phi \sin \phi \cos \theta, \cos \phi \sin \phi \sin \theta, \cos \phi \cos \phi)ψ(θ,ϕ)=(cosϕsinϕcosθ,cosϕsinϕsinθ,cosϕcosϕ)
  5. ϕ ( θ , z ) = ( cos 2 θ , sin θ cos θ , z ) ϕ ( θ , z ) = cos 2 θ , sin θ cos θ , z phi(theta,z)=(cos^(2)theta,sin theta cos theta,z)\phi(\theta, z)=\left(\cos ^{2} \theta, \sin \theta \cos \theta, z\right)ϕ(θ,z)=(cos2θ,sinθcosθ,z)
  6. ϕ ( r , θ ) = ( r cos θ , r sin θ , r 2 1 ) ϕ ( r , θ ) = r cos θ , r sin θ , r 2 1 phi(r,theta)=(r cos theta,r sin theta,sqrt(r^(2)-1))\phi(r, \theta)=\left(r \cos \theta, r \sin \theta, \sqrt{r^{2}-1}\right)ϕ(r,θ)=(rcosθ,rsinθ,r21)
  7. ϕ ( r , θ ) = ( r cos θ , r sin θ , r 2 + 1 ) ϕ ( r , θ ) = r cos θ , r sin θ , r 2 + 1 phi(r,theta)=(r cos theta,r sin theta,sqrt(r^(2)+1))\phi(r, \theta)=\left(r \cos \theta, r \sin \theta, \sqrt{r^{2}+1}\right)ϕ(r,θ)=(rcosθ,rsinθ,r2+1)
  8. ϕ ( θ , z ) = ( θ cos θ , θ sin θ , z ) ϕ ( θ , z ) = ( θ cos θ , θ sin θ , z ) phi(theta,z)=(theta cos theta,theta sin theta,z)\phi(\theta, z)=(\theta \cos \theta, \theta \sin \theta, z)ϕ(θ,z)=(θcosθ,θsinθ,z)
    2.11 φ ( x , y ) = ( x , y , f ( x , y ) ) 2.11 φ ( x , y ) = ( x , y , f ( x , y ) ) 2.11 varphi(x,y)=(x,y,f(x,y))2.11 \varphi(x, y)=(x, y, f(x, y))2.11φ(x,y)=(x,y,f(x,y))
    2.13 Ψ ( θ , φ ) = ( 2 sin φ cos θ , 2 sin φ sin θ , 2 cos φ ) , 0 θ 2 π 2.13 Ψ ( θ , φ ) = ( 2 sin φ cos θ , 2 sin φ sin θ , 2 cos φ ) , 0 θ 2 π 2.13 Psi(theta,varphi)=(2sin varphi cos theta,2sin varphi sin theta,2cos varphi),0 <= theta <= 2pi2.13 \Psi(\theta, \varphi)=(2 \sin \varphi \cos \theta, 2 \sin \varphi \sin \theta, 2 \cos \varphi), 0 \leq \theta \leq 2 \pi2.13Ψ(θ,φ)=(2sinφcosθ,2sinφsinθ,2cosφ),0θ2π,
    0 θ 2 π 0 θ 2 π 0 <= theta <= 2pi0 \leq \theta \leq 2 \pi0θ2π, π 4 ϕ π 2 π 4 ϕ π 2 (pi)/(4) <= phi <= (pi)/(2)\frac{\pi}{4} \leq \phi \leq \frac{\pi}{2}π4ϕπ2
    2.15 2 , 0 , 4 , 0 , 3 , 2 2.15 2 , 0 , 4 , 0 , 3 , 2 2.15(:2,0,4:),(:0,3,2:)2.15\langle 2,0,4\rangle,\langle 0,3,2\rangle2.152,0,4,0,3,2
    2.17

1. The x x xxx-axis

  1. The z z zzz-axis
  2. The line y = z y = z y=zy=zy=z and x = 0 x = 0 x=0x=0x=0
  3. The line x = y = z x = y = z x=y=zx=y=zx=y=z
    2.18 ϕ ( t ) = ( t , 1 t , 1 2 t + 2 t 2 ) 2.18 ϕ ( t ) = t , 1 t , 1 2 t + 2 t 2 2.18 phi(t)=(t,1-t,sqrt(1-2t+2t^(2)))2.18 \phi(t)=\left(t, 1-t, \sqrt{1-2 t+2 t^{2}}\right)2.18ϕ(t)=(t,1t,12t+2t2)
    2.19
  4. ϕ ( θ ) = ( 2 cos θ , 2 sin θ , 4 ) , 0 θ 2 π ϕ ( θ ) = ( 2 cos θ , 2 sin θ , 4 ) , 0 θ 2 π phi(theta)=(2cos theta,2sin theta,4),0 <= theta <= 2pi\phi(\theta)=(2 \cos \theta, 2 \sin \theta, 4), 0 \leq \theta \leq 2 \piϕ(θ)=(2cosθ,2sinθ,4),0θ2π
  5. ψ ( t ) = ( t , ± 4 t 2 , 4 ) , 2 t 2 ψ ( t ) = t , ± 4 t 2 , 4 , 2 t 2 psi(t)=(t,+-sqrt(4-t^(2)),4),-2 <= t <= 2\psi(t)=\left(t, \pm \sqrt{4-t^{2}}, 4\right),-2 \leq t \leq 2ψ(t)=(t,±4t2,4),2t2

2.20

  1. Ψ ( ρ , θ , φ ) = ( ρ sin φ cos θ , ρ sin φ sin θ , ρ cos φ ) Ψ ( ρ , θ , φ ) = ( ρ sin φ cos θ , ρ sin φ sin θ , ρ cos φ ) Psi(rho,theta,varphi)=(rho sin varphi cos theta,rho sin varphi sin theta,rho cos varphi)\Psi(\rho, \theta, \varphi)=(\rho \sin \varphi \cos \theta, \rho \sin \varphi \sin \theta, \rho \cos \varphi)Ψ(ρ,θ,φ)=(ρsinφcosθ,ρsinφsinθ,ρcosφ), 0 ρ 1 , 0 θ 2 π , 0 φ π 0 ρ 1 , 0 θ 2 π , 0 φ π 0 <= rho <= 1,0 <= theta <= 2pi,0 <= varphi <= pi0 \leq \rho \leq 1,0 \leq \theta \leq 2 \pi, 0 \leq \varphi \leq \pi0ρ1,0θ2π,0φπ
  2. Ψ ( ρ , θ , φ ) = ( ρ sin φ cos θ , ρ sin φ sin θ , ρ cos φ ) , 0 Ψ ( ρ , θ , φ ) = ( ρ sin φ cos θ , ρ sin φ sin θ , ρ cos φ ) , 0 Psi(rho,theta,varphi)=(rho sin varphi cos theta,rho sin varphi sin theta,rho cos varphi),0 <=\Psi(\rho, \theta, \varphi)=(\rho \sin \varphi \cos \theta, \rho \sin \varphi \sin \theta, \rho \cos \varphi), 0 \leqΨ(ρ,θ,φ)=(ρsinφcosθ,ρsinφsinθ,ρcosφ),0 ρ 1 , 0 θ π 2 , 0 ϕ π 2 ρ 1 , 0 θ π 2 , 0 ϕ π 2 rho <= 1,0 <= theta <= (pi)/(2),0 <= phi <= (pi)/(2)\rho \leq 1,0 \leq \theta \leq \frac{\pi}{2}, 0 \leq \phi \leq \frac{\pi}{2}ρ1,0θπ2,0ϕπ2

2.21

1. ϕ ( r , θ , z ) = ( r cos θ , r sin θ , z ) , 0 r 1 , 0 θ 2 π , 0 z 1 2. ϕ ( r , θ , z ) = ( r cos θ , r sin θ , z ) , 1 r 2 , 0 θ 2 π , 0 z 2  1.  ϕ ( r , θ , z ) = ( r cos θ , r sin θ , z ) , 0 r 1 , 0 θ 2 π , 0 z 1  2.  ϕ ( r , θ , z ) = ( r cos θ , r sin θ , z ) , 1 r 2 , 0 θ 2 π , 0 z 2 {:[" 1. "phi(r","theta","z)=(r cos theta","r sin theta","z)","0 <= r <= 1","0 <= theta <= 2pi","0 <= z <= 1],[" 2. "phi(r","theta","z)=(r cos theta","r sin theta","z)","1 <= r <= 2","0 <= theta <= 2pi","0 <= z <= 2]:}\begin{aligned} & \text { 1. } \phi(r, \theta, z)=(r \cos \theta, r \sin \theta, z), 0 \leq r \leq 1,0 \leq \theta \leq 2 \pi, 0 \leq z \leq 1 \\ & \text { 2. } \phi(r, \theta, z)=(r \cos \theta, r \sin \theta, z), 1 \leq r \leq 2,0 \leq \theta \leq 2 \pi, 0 \leq z \leq 2 \end{aligned} 1. ϕ(r,θ,z)=(rcosθ,rsinθ,z),0r1,0θ2π,0z1 2. ϕ(r,θ,z)=(rcosθ,rsinθ,z),1r2,0θ2π,0z2

2.22

ϕ ( t , θ ) = ( [ t f 2 ( θ ) + ( 1 t ) f 1 ( θ ) ] cos θ , [ t f 2 ( θ ) + ( 1 t ) f 1 ( θ ) ] sin θ ) ϕ ( t , θ ) = t f 2 ( θ ) + ( 1 t ) f 1 ( θ ) cos θ , t f 2 ( θ ) + ( 1 t ) f 1 ( θ ) sin θ phi(t,theta)=([tf_(2)(theta)+(1-t)f_(1)(theta)]cos theta,[tf_(2)(theta)+(1-t)f_(1)(theta)]sin theta)\phi(t, \theta)=\left(\left[t f_{2}(\theta)+(1-t) f_{1}(\theta)\right] \cos \theta,\left[t f_{2}(\theta)+(1-t) f_{1}(\theta)\right] \sin \theta\right)ϕ(t,θ)=([tf2(θ)+(1t)f1(θ)]cosθ,[tf2(θ)+(1t)f1(θ)]sinθ)
0 t 1 , a θ b 0 t 1 , a θ b 0 <= t <= 1,a <= theta <= b0 \leq t \leq 1, a \leq \theta \leq b0t1,aθb
2.23 φ ( r , θ ) = ( 3 r cos θ , 2 r sin θ ) , 0 r 1 , 0 θ 2 π 2.23 φ ( r , θ ) = ( 3 r cos θ , 2 r sin θ ) , 0 r 1 , 0 θ 2 π 2.23 varphi(r,theta)=(3r cos theta,2r sin theta),0 <= r <= 1,0 <= theta <= 2pi2.23 \varphi(r, \theta)=(3 r \cos \theta, 2 r \sin \theta), 0 \leq r \leq 1,0 \leq \theta \leq 2 \pi2.23φ(r,θ)=(3rcosθ,2rsinθ),0r1,0θ2π

Chapter 4

4.2

  1. 1 , 4 , 10 1 , 4 , 10 -1,4,10-1,4,101,4,10
  2. d y = 4 d x d y = 4 d x dy=-4dxd y=-4 d xdy=4dx

4.3

  1. 3 d x 3 d x 3dx3 d x3dx
  2. 1 2 d y 1 2 d y (1)/(2)dy\frac{1}{2} d y12dy
  3. 3 d x + 1 2 d y 3 d x + 1 2 d y 3dx+(1)/(2)dy3 d x+\frac{1}{2} d y3dx+12dy
  4. 8 d x + 6 d y 8 d x + 6 d y 8dx+6dy8 d x+6 d y8dx+6dy

4.5

  1. ω ( V 1 ) = 8 , v ( V 1 ) = 1 , ω ( V 2 ) = 1 ω V 1 = 8 , v V 1 = 1 , ω V 2 = 1 omega(V_(1))=-8,v(V_(1))=1,omega(V_(2))=-1\omega\left(V_{1}\right)=-8, v\left(V_{1}\right)=1, \omega\left(V_{2}\right)=-1ω(V1)=8,v(V1)=1,ω(V2)=1 and v ( V 2 ) = 2 v V 2 = 2 v(V_(2))=2v\left(V_{2}\right)=2v(V2)=2.
  2. -15
  3. 5
    4.15-127
    4.16 c 1 = 11 , c 2 = 4 4.16 c 1 = 11 , c 2 = 4 4.16c_(1)=-11,c_(2)=44.16 c_{1}=-11, c_{2}=44.16c1=11,c2=4, and c 3 = 3 c 3 = 3 c_(3)=3c_{3}=3c3=3
    4.17
  4. 2 d x d y 2 d x d y 2dx^^dy2 d x \wedge d y2dxdy
  5. d x ( d y + d z ) d x ( d y + d z ) dx^^(dy+dz)d x \wedge(d y+d z)dx(dy+dz)
  6. d x ( 2 d y + d z ) d x ( 2 d y + d z ) dx^^(2dy+dz)d x \wedge(2 d y+d z)dx(2dy+dz)
  7. ( d x + 3 d z ) ( d y + 4 d z ) ( d x + 3 d z ) ( d y + 4 d z ) (dx+3dz)^^(dy+4dz)(d x+3 d z) \wedge(d y+4 d z)(dx+3dz)(dy+4dz)
    4.29252
    4.30
  1. -87
  2. -29
  3. 5
    4.31 d x d y d z 4.31 d x d y d z 4.31 dx^^dy^^dz4.31 d x \wedge d y \wedge d z4.31dxdydz 4.33
  1. z ( x y ) d z d x + z ( x + y ) d z d y z ( x y ) d z d x + z ( x + y ) d z d y z(x-y)dz^^dx+z(x+y)dz^^dyz(x-y) d z \wedge d x+z(x+y) d z \wedge d yz(xy)dzdx+z(x+y)dzdy
  2. 4 d x d y d z 4 d x d y d z -4dx^^dy^^dz-4 d x \wedge d y \wedge d z4dxdydz

Chapter 5

5.1

  1. 2 , 3 , 1 , 2 , 3 , 2 2 , 3 , 1 , 2 , 3 , 2 (:2,3,1:),(:2,3,2:)\langle 2,3,1\rangle,\langle 2,3,2\rangle2,3,1,2,3,2
  2. 6 d x d y + 3 d y d z 2 d x d z 6 d x d y + 3 d y d z 2 d x d z 6dx^^dy+3dy^^dz-2dx^^dz6 d x \wedge d y+3 d y \wedge d z-2 d x \wedge d z6dxdy+3dydz2dxdz
  3. 5
  4. x 2 y z 2 x 5 z 2 y 3 + x 3 y 2 x 2 y z 2 x 5 z 2 y 3 + x 3 y 2 x^(2)yz^(2)-x^(5)z^(2)-y^(3)+x^(3)y^(2)x^{2} y z^{2}-x^{5} z^{2}-y^{3}+x^{3} y^{2}x2yz2x5z2y3+x3y2
    5.2 1 6 5.2 1 6 5.2(1)/(6)5.2 \frac{1}{6}5.216
    5.3 ± 4 π 5.3 ± 4 π 5.3+-4pi5.3 \pm 4 \pi5.3±4π. (You don't have enough information yet to properly tell the sign.)
    5.4 4п
    5.5 1 3 5.6 5.5 1 3 5.6 5.5(1)/(3)5.65.5 \frac{1}{3} 5.65.5135.6
1. 17 12 2. 29 6  1.  17 12  2.  29 6 {:[" 1. "-(17)/(12)],[" 2. "-(29)/(6)]:}\begin{aligned} & \text { 1. }-\frac{17}{12} \\ & \text { 2. }-\frac{29}{6} \end{aligned} 1. 1712 2. 296
5.8 3п/5
5.9 1 6 5.9 1 6 5.9(1)/(6)5.9 \frac{1}{6}5.916
5.10
  1. Opposite orientation
  2. Same orientation
  3. Does not determine an orientation
    5.1116
    5.1216
    5.13 1 5 5.13 1 5 5.13(1)/(5)5.13 \frac{1}{5}5.1315
    5.14 2 3 cos 6 3 2 sin 4 2 3 5.14 2 3 cos 6 3 2 sin 4 2 3 5.14(2)/(3)cos 6-(3)/(2)sin 4-(2)/(3)5.14 \frac{2}{3} \cos 6-\frac{3}{2} \sin 4-\frac{2}{3}5.1423cos632sin423
    5.15 1 6 5.15 1 6 5.15(1)/(6)5.15 \frac{1}{6}5.1516
    5.164
    5.17 32 3 5.17 32 3 5.17(32)/(3)5.17 \frac{32}{3}5.17323
    5.192
    5.20 6п
    5.22 14 3 5.22 14 3 5.22(14)/(3)5.22 \frac{14}{3}5.22143
    5.23 7 π 6 5.23 7 π 6 5.23(-7pi)/(6)5.23 \frac{-7 \pi}{6}5.237π6
    5.24 π 3 ( 2 3 / 2 1 ) 5.24 π 3 2 3 / 2 1 5.24(pi)/(3)(2^(3//2)-1)5.24 \frac{\pi}{3}\left(2^{3 / 2}-1\right)5.24π3(23/21)
    5.25 2п

Chapter 6

6.1

1. V ω ( W ) = 62 , W ω ( V ) = 4 2. 66  1.  V ω ( W ) = 62 , W ω ( V ) = 4  2.  66 {:[" 1. "grad_(V)omega(W)=-62","grad_(W)omega(V)=4],[" 2. "-66]:}\begin{aligned} & \text { 1. } \nabla_{V} \omega(W)=-62, \nabla_{W} \omega(V)=4 \\ & \text { 2. }-66 \end{aligned} 1. Vω(W)=62,Wω(V)=4 2. 66
6.3 a ω = ( 2 x 1 ) d x d y 6.3 a ω = ( 2 x 1 ) d x d y 6.3 a omega=(-2x-1)dx^^dy6.3 a \omega=(-2 x-1) d x \wedge d y6.3aω=(2x1)dxdy
6.6 d ( x 2 y d x d y + y 2 z d y d z ) = 0 6.6 d x 2 y d x d y + y 2 z d y d z = 0 6.6 d(x^(2)ydx^^dy+y^(2)zdy^^dz)=06.6 d\left(x^{2} y d x \wedge d y+y^{2} z d y \wedge d z\right)=06.6d(x2ydxdy+y2zdydz)=0
6.7 1 , 1 , 1 6.7 1 , 1 , 1 6.7-1,1,16.7-1,1,16.71,1,1

6.11

  1. ( sin x cos y ) d x d y ( sin x cos y ) d x d y (-sin x-cos y)dx^^dy(-\sin x-\cos y) d x \wedge d y(sinxcosy)dxdy
  2. ( 3 x 2 z 2 x y ) d x d y ( x 3 + 1 ) d y d z 3 x 2 z 2 x y d x d y x 3 + 1 d y d z (3x^(2)z-2xy)dx^^dy-(x^(3)+1)dy^^dz\left(3 x^{2} z-2 x y\right) d x \wedge d y-\left(x^{3}+1\right) d y \wedge d z(3x2z2xy)dxdy(x3+1)dydz
  3. ( y 2 9 z 8 ) d x d y d z y 2 9 z 8 d x d y d z (y^(2)-9z^(8))dx^^dy^^dz\left(y^{2}-9 z^{8}\right) d x \wedge d y \wedge d z(y29z8)dxdydz
  4. 0
    6.12 ( 3 x 4 y 2 4 x y 6 z ) d x d y d z 6.12 3 x 4 y 2 4 x y 6 z d x d y d z 6.12(3x^(4)y^(2)-4xy^(6)z)dx^^dy^^dz6.12\left(3 x^{4} y^{2}-4 x y^{6} z\right) d x \wedge d y \wedge d z6.12(3x4y24xy6z)dxdydz
    6.14
  5. x d y x d y xdyx d yxdy
  6. x d y d z x d y d z xdy^^dzx d y \wedge d zxdydz
  7. x y z x y z xyzx y zxyz
  8. x y 2 z 2 x y 2 z 2 xy^(2)z^(2)x y^{2} z^{2}xy2z2
  9. sin ( x y 2 ) d x + sin ( x y 2 ) d y sin x y 2 d x + sin x y 2 d y sin(xy^(2))dx+sin(xy^(2))dy\sin \left(x y^{2}\right) d x+\sin \left(x y^{2}\right) d ysin(xy2)dx+sin(xy2)dy

Chapter 7

7.1
  1. 2 , 2 2 , 2 2,-22,-22,2
  2. 1
    7.121
    7.1327
    7.14 35 3 7.14 35 3 7.14(35)/(3)7.14 \frac{35}{3}7.14353
    7.15
  3. 0
  4. C 1 ω = C 2 ω = 1 3 C 1 ω = C 2 ω = 1 3 int_(C_(1))omega=int_(C_(2))omega=(1)/(3)\int_{C_{1}} \omega=\int_{C_{2}} \omega=\frac{1}{3}C1ω=C2ω=13
  5. 2 3 2 3 -(2)/(3)-\frac{2}{3}23

7.16

  1. 0
  2. 0
  3. If L , R , T L , R , T L,R,TL, R, TL,R,T, and B B BBB represent the 1 -cells that are the left, right, top and bottom of Q Q QQQ, then
Q ω = ( R L ) ( T B ) ω = R ω L ω T ω + B ω = 24 0 28 1 2 + 4 1 2 = 0 Q ω = ( R L ) ( T B ) ω = R ω L ω T ω + B ω = 24 0 28 1 2 + 4 1 2 = 0 int_(del Q)omega=int_((R-L)-(T-B))omega=int_(R)omega-int_(L)omega-int_(T)omega+int_(B)omega=24-0-28(1)/(2)+4(1)/(2)=0\int_{\partial Q} \omega=\int_{(R-L)-(T-B)} \omega=\int_{R} \omega-\int_{L} \omega-\int_{T} \omega+\int_{B} \omega=24-0-28 \frac{1}{2}+4 \frac{1}{2}=0Qω=(RL)(TB)ω=RωLωTω+Bω=2402812+412=0
  1. Opposite
  2. 4 1 2 4 1 2 4(1)/(2)4 \frac{1}{2}412
  3. 4 1 2 4 1 2 4(1)/(2)4 \frac{1}{2}412
    7.19 8 7.20 6п
    7.21 2 3 ( e 1 e ) 7.21 2 3 e 1 e 7.21^((2)/(3))(e-(1)/(e))7.21^{\frac{2}{3}}\left(e-\frac{1}{e}\right)7.2123(e1e)
    7.240
    7.25 4 3 7.25 4 3 7.25^((4)/(3))7.25^{\frac{4}{3}}7.2543

Chapter 9

9.9 a T = 2 d x d y 9.9 a T = 2 d x d y 9.9a^(')T^(')=2dx^^dy9.9 a^{\prime} T^{\prime}=2 d x \wedge d y9.9aT=2dxdy and a t = 0 a t = 0 at=0a t=0at=0

1

Figure drawn by Stephan Schoenenberger. Taken from Introductory Lectures on Contact Geometry by John B. Etnyre.